Minor surgery on horses under field conditions requires a reliable short-term anesthetic producing good muscle relaxation, minimal cardiovascular and respiratory changes, and a quick, smooth recovery with minimal aftereffects so that horses can be left unattended. An article reports that for a sample of n = 69 horses to which ketamine was administered under certain conditions, the sample average lateral recumbency (lying-down) time was 18.94 min and the standard deviation was 8.3 min. Does this data suggest that true average lateral recumbency time under these conditions is less than 20 min? Test the appropriate hypotheses at level of significance 0.10.State the appropriate null and alternative hypotheses.

Answers

Answer 1

Answer:

Null hypothesis:[tex]\mu \geq 20[/tex]        

Alternative hypothesis:[tex]\mu < 20[/tex]

[tex]P(t_{68}<-1.06)=0.146[/tex]

If we compare the p value and the significance level given [tex]\alpha=0.1[/tex] we see that [tex]p_v>\alpha[/tex] so we can conclude that we FAIL reject the null hypothesis, and the the actual sample average lateral recumbency is not significantly lower than 20.        

Step-by-step explanation:

1) Data given and notation        

[tex]\bar X=18.94[/tex] represent the mean for the sample    

[tex]s=8.3[/tex] represent the standard deviation for the sample        

[tex]n=69[/tex] sample size        

[tex]\mu_o =20[/tex] represent the value that we want to test      

[tex]\alpha[/tex] represent the significance level for the hypothesis test.      

t would represent the statistic (variable of interest)        

[tex]p_v[/tex] represent the p value for the test (variable of interest)    

2) State the null and alternative hypotheses.        

We need to conduct a hypothesis in order to determine if the true average lateral recumbency time under these conditions is less than 20 min:      

Null hypothesis:[tex]\mu \geq 20[/tex]        

Alternative hypothesis:[tex]\mu < 20[/tex]        

We don't know the population deviation, so for this case is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:        

[tex]t=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}}[/tex] (1)        

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".    

3) Calculate the statistic        

We can replace in formula (1) the info given like this:        

[tex]t=\frac{18.94-20}{\frac{8.3}{\sqrt{69}}}=-1.06[/tex]        

4) Calculate the P-value        

First we need to calculate the degrees of freedom

[tex]df=n-1=69-1=68[/tex]

The critical value for this case would be :

[tex]P(t_{68}<-1.06)=0.146[/tex]

5) Conclusion      

If we compare the p value and the significance level given [tex]\alpha=0.1[/tex] we see that [tex]p_v>\alpha[/tex] so we can conclude that we FAIL reject the null hypothesis, and the the actual sample average lateral recumbency is not significantly lower than 20.        


Related Questions

Demand for your tie-dyed T-shirts is given by the formula q = 510 − 90p0.5 where q is the number of T-shirts you can sell each month at a price of p dollars. If you currently sell T-shirts for $15 each and you raise your price by $2 per month, how fast will the demand drop? (Round your answer to the nearest whole number.)

Answers

Final answer:

The demand for tie-dyed T-shirts will drop by approximately 23 T-shirts when the price is increased from $15 to $17. This calculation is based on the function q = 510 − 90[tex]p^{0.5}[/tex], which represents the demand in relation to the price.

Explanation:

The student is asking how fast the demand for tie-dyed T-shirts will drop if the price increases by $2 when the demand is represented by the function q = 510 − 90[tex]p^{0.5}[/tex]. Since the current price is $15, we need to calculate the demand at both $15 and $17 to find the rate of change in demand when the price is raised by $2.

To find the demand at the current price of $15, substitute p = 15 into the demand function:
q = 510 − 90[tex](15)^{0.5}[/tex]
q = 510 − 90(3.87)
q = 510 − 348.3
q = 161.7

To find the demand when the price is $17, substitute p = 17 into the demand function:
q = 510 − 90[tex](17)^{0.5}[/tex]
q = 510 − 90(4.12)
q = 510 − 371
q = 139

The change in demand is the difference between the two quantities:
Δq = 161.7 - 139 = 22.7
The demand drops by approximately 23 T-shirts when the price increases by $2.

The demand will drop by approximately [tex]\( \boed{23} \)[/tex] T-shirts per month.

Step 1

To determine how fast the demand for tie-dyed T-shirts will drop when the price increases by $2, we need to calculate the rate of change of the demand q with respect to the price p. The given demand function is:

[tex]\[ q = 510 - 90p^{0.5} \][/tex]

We need to find the derivative of q with respect to p, which gives us the rate of change of demand with respect to price.

Step 2

First, let's find the derivative [tex]\( \frac{dq}{dp} \):[/tex]

[tex]\[ q = 510 - 90p^{0.5} \]\[ \frac{dq}{dp} = -90 \cdot \frac{1}{2} p^{-0.5} \]\[ \frac{dq}{dp} = -45 p^{-0.5} \]\[ \frac{dq}{dp} = -45 \cdot \frac{1}{\sqrt{p}} \][/tex]

Step 3

Now, we need to evaluate this derivative at the current price [tex]\( p = 15 \)[/tex]:

[tex]\[ \frac{dq}{dp} \bigg|_{p=15} = -45 \cdot \frac{1}{\sqrt{15}} \][/tex]

Calculate \[tex]( \sqrt{15} \)[/tex]:

[tex]\[ \sqrt{15} \approx 3.872 \][/tex]

So,

[tex]\[ \frac{dq}{dp} \bigg|_{p=15} = -45 \cdot \frac{1}{3.872} \approx -11.62 \][/tex]

This means the rate of change of demand with respect to price at [tex]\( p = 15 \)[/tex] is approximately [tex]\(-11.62\)[/tex] T-shirts per dollar.

Given that the price is increased by $2, we need to find how much the demand drops for this price increase:

[tex]\[ \frac{dq}{dp} \times 2 \approx -11.62 \times 2 \approx |-23.24| \][/tex]

Rounded to the nearest whole number, the demand drops by approximately 23 T-shirts per month when the price is raised by $2.

Does second-hand smoke increase the risk of a low birthweight? A baby is considered having low birth weight if he/she weighs less than 5.5 pounds at birth. According to the National Center of Health Statistics, about 7.8% of all babies born in the U.S. are categorized as low birthweight.
Suspecting that the national percentage is higher than 7.8%, researchers randomly select 1200 babies whose mothers had extensive exposure to second-hand smoke during pregnancy and find that 10.4% of the sampled babies are categorized as low birth weight.
Let p be the proportion of all babies in the U.S. that are categorized as "low birth weight." Give the null and alternative hypotheses for this research question.
H 0 : p = 0.078
H 0 : p = 0.078
H a : p > 0.078
H 0 : p = 0.104

Answers

Final answer:

Researchers are testing the hypothesis that second-hand smoke increases the risk of low birthweight, with a null hypothesis of H0: p = 0.078, and an alternative hypothesis of Ha: p > 0.078.

Explanation:

The researchers are questioning whether second-hand smoke increases the risk of low birthweight in babies. They have a sample where 10.4% of babies born to mothers with extensive exposure to second-hand smoke are categorized as having low birth weight. This suggests the proportion might be higher than the national average of 7.8%. The null hypothesis (H0), which is the hypothesis to be tested, states that the proportion of low birth weight for the population is equal to the national average, thus H0: p = 0.078. The alternative hypothesis (Ha), representing the researchers' suspicion, is that the proportion is greater than the national average, hence Ha: p > 0.078. The last hypothesis option provided, H0: p = 0.104, is not relevant for the null hypothesis in this context since this value represents the sample proportion, not the population proportion they are testing against.

While her husband spent 2½ hours picking out new speakers, a statistician decided to determine whether the percent of men who enjoy shopping for electronic equipment is higher than the percent of women who enjoy shopping for electronic equipment. The population was Saturday afternoon shoppers. Out of 66 men, 23 said they enjoyed the activity. Eight of the 23 women surveyed claimed to enjoy the activity. Interpret the results of the survey. Conduct a hypothesis test at the 5% level. Let the subscript m = men and w = women.

State the distribution to use for the test. (Round your answers to four decimal places.)

Answers

Answer:

[tex]z=\frac{p_{M}-p_{W}}{\sqrt{\hat p (1-\hat p)(\frac{1}{n_{M}}+\frac{1}{n_{W}})}}[/tex]   (1)

[tex]z=\frac{0.34848-0.34782}{\sqrt{0.34831(1-0.34831)(\frac{1}{66}+\frac{1}{23})}}=0.0057[/tex]  

[tex]p_v =P(Z>0.0057)=0.4977[/tex]  

The p value is a very high value and using the significance level [tex]\alpha=0.05[/tex] we see that [tex]p_v>\alpha[/tex] so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can say the the percent of men who enjoy shopping for electronic equipment is NOT significantly higher than the percent of women who enjoy shopping for electronic equipment.  

Step-by-step explanation:

1) Data given and notation  

[tex]X_{M}=23[/tex] represent the number of men that said they enjoyed the activity of Saturday afternoon shopping

[tex]X_{W}=8[/tex] represent the number of women that said they enjoyed the activity of Saturday afternoon shopping

[tex]n_{M}=66[/tex] sample of male selected

[tex]n_{W}=23[/tex] sample of demale selected

[tex]p_{M}=\frac{23}{66}=0.34848[/tex] represent the proportion of men that said they enjoyed the activity of Saturday afternoon shopping

[tex]p_{W}=\frac{8}{23}=0.34782[/tex] represent the proportion of women with red/green color blindness  

z would represent the statistic (variable of interest)  

[tex]p_v[/tex] represent the value for the test (variable of interest)

2) Concepts and formulas to use  

We need to conduct a hypothesis in order to check if the percent of men who enjoy shopping for electronic equipment is higher than the percent of women who enjoy shopping for electronic equipment , the system of hypothesis would be:  

Null hypothesis:[tex]p_{M} \leq p_{W}[/tex]  

Alternative hypothesis:[tex]p_{M} > p_{W}[/tex]  

We need to apply a z test to compare proportions, and the statistic is given by:  

[tex]z=\frac{p_{M}-p_{W}}{\sqrt{\hat p (1-\hat p)(\frac{1}{n_{M}}+\frac{1}{n_{W}})}}[/tex]   (1)

Where [tex]\hat p=\frac{X_{M}+X_{W}}{n_{M}+n_{W}}=\frac{23+8}{66+23}=0.34831[/tex]

3) Calculate the statistic

Replacing in formula (1) the values obtained we got this:  

[tex]z=\frac{0.34848-0.34782}{\sqrt{0.34831(1-0.34831)(\frac{1}{66}+\frac{1}{23})}}=0.0057[/tex]  

4) Statistical decision

Using the significance level provided [tex]\alpha=0.05[/tex], the next step would be calculate the p value for this test.  

Since is a one side right tail test the p value would be:  

[tex]p_v =P(Z>0.0057)=0.4977[/tex]  

So the p value is a very high value and using the significance level [tex]\alpha=0.05[/tex] we see that [tex]p_v>\alpha[/tex] so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can say the the percent of men who enjoy shopping for electronic equipment is NOT significantly higher than the percent of women who enjoy shopping for electronic equipment.  

A two-proportion z-test shows there is insufficient evidence at the 5% significance level to conclude that a higher proportion of men enjoy shopping for electronic equipment compared to women. The p-value is greater than the alpha value, leading to not rejecting the null hypothesis.

A hypothesis test can be used to determine if the percent of men who enjoy shopping for electronic equipment is higher than the percent of women who enjoy the same activity. We will perform a two-proportion z-test for this purpose. The hypotheses are as follows:

Null Hypothesis (H0): pm = pw

Alternative Hypothesis (Ha): pm > pw

Here, pm is the proportion of men who enjoy shopping for electronic equipment, and pw is the proportion of women who enjoy shopping for electronic equipment.

Given:

Out of 66 men, 23 enjoy shopping (pm = 23/66 ≈ 0.3485).Out of 23 women, 8 enjoy shopping (pw = 8/23 ≈ 0.3478).

The test statistic for the difference in proportions is calculated as:

z = (pm - pw) / [tex]\sqrt{(p*(1 - p*)(1/nm + 1/nw))}[/tex]

Where p* = (xm + xw) / (nm + nw)

Substituting the values:

p* = (23 + 8) / (66 + 23) ≈ 0.3484z = (0.3485 - 0.3478) / [tex]\sqrt{(0.3484 * (1 - 0.3484) * (1/66 + 1/23))}[/tex] ≈ 0.0130

The calculated z-value is approximately 0.0130.

The corresponding p-value for a z-value of 0.0130 is approximately 0.4948. Since p-value > alpha (0.4948 > 0.05), we do not reject the null hypothesis.

Conclusion:

At the 5% significance level, there is insufficient evidence to conclude that the proportion of men who enjoy shopping for electronic equipment is significantly higher than that of women.

Find angle x on the given figure

Answers

Answer:

41 is correct

Step-by-step explanation:

90 + 49 = 139

180 - 139 = 41

Find the t-coordinates of all critical points of the given function. Determine whether each critical point is a relative maximum, minimum, or neither by first applying the second derivative test, and, if the test fails, by some other method.

f(t)= -3t^3+2t

a.) f has (a relative maximum, relative minimum, or no relative extrema) at the critical point t=________ * (smaller t-value)

b.) f has (a relative maximum, relative minimum, or no relative extrema) at the crtical point t=__________* (larger t-value)

Answers

Answer:

f has relative maximum at t = [tex]+\frac{\sqrt2}{3}[/tex]

and

f has relative minimum at t = [tex]-\frac{\sqrt2}{3}[/tex]

Step-by-step explanation:

Data provided in the question:

f(t) = -3t³ + 2t

Now,

To find the  points of maxima or minima, differentiating with respect to t and putting it equals to zero

thus,

f'(t) = (3)(-3t²) + 2 = 0

or

-9t² + 2 = 0

or

t² = [tex]\frac{2}{9}[/tex]

or

t = [tex]\pm\frac{\sqrt2}{3}[/tex]

to check for maxima or minima, again differentiating with respect to t

f''(t) = 2(-9t) + 0 = -18t

substituting the value of t

at t = [tex]+\frac{\sqrt2}{3}[/tex]

f''(t) =  [tex](-18)\times\frac{\sqrt2}{3}[/tex]

=  - 6√2 < 0 i.e maxima

and at  t = [tex]-\frac{\sqrt2}{3}[/tex]

f''(t) =  [tex](-18)\times(-\frac{\sqrt2}{3})[/tex]  

= 6√2 > 0 i.e minima

Hence,

f has relative maximum at t = [tex]+\frac{\sqrt2}{3}[/tex]

and

f has relative minimum at t = [tex]-\frac{\sqrt2}{3}[/tex]

The joint probability density function of X and Y is given by fX,Y (x, y) = ( 6 7 x 2 + xy 2 if 0 < x < 1, 0 < y < 2 0 otherwise. (a) Verify that this is indeed a joint density function. (b) Compute the density function of X. (c) Find P(X > Y ). (d) Find P(Y > 1/2 | X < 1/2). (e) Find E(X). (f) Find E(Y

Answers

I'm going to assume the joint density function is

[tex]f_{X,Y}(x,y)=\begin{cases}\frac67(x^2+\frac{xy}2\right)&\text{for }0<x<1,0<y<2\\0&\text{otherwise}\end{cases}[/tex]

a. In order for [tex]f_{X,Y}[/tex] to be a proper probability density function, the integral over its support must be 1.

[tex]\displaystyle\int_0^2\int_0^1\frac67\left(x^2+\frac{xy}2\right)\,\mathrm dx\,\mathrm dy=\frac67\int_0^2\left(\frac13+\frac y4\right)\,\mathrm dy=1[/tex]



b. You get the marginal density [tex]f_X[/tex] by integrating the joint density over all possible values of [tex]Y[/tex]:

[tex]f_X(x)=\displaystyle\int_0^2f_{X,Y}(x,y)\,\mathrm dy=\boxed{\begin{cases}\frac67(2x^2+x)&\text{for }0<x<1\\0&\text{otherwise}\end{cases}}[/tex]

c. We have

[tex]P(X>Y)=\displaystyle\int_0^1\int_0^xf_{X,Y}(x,y)\,\mathrm dy\,\mathrm dx=\int_0^1\frac{15}{14}x^3\,\mathrm dx=\boxed{\frac{15}{56}}[/tex]

d. We have

[tex]\displaystyle P\left(X<\frac12\right)=\int_0^2\int_0^{1/2}f_{X,Y}(x,y)\,\mathrm dx\,\mathrm dy=\frac5{28}[/tex]

and by definition of conditional probability,

[tex]P\left(Y>\dfrac12\mid X<\dfrac12\right)=\dfrac{P\left(Y>\frac12\text{ and }X<\frac12\right)}{P\left(X<\frac12\right)}[/tex]

[tex]\displaystyle=\dfrac{28}5\int_{1/2}^2\int_0^{1/2}f_{X,Y}(x,y)\,\mathrm dx\,\mathrm dy=\boxed{\frac{69}{80}}[/tex]

e. We can find the expectation of [tex]X[/tex] using the marginal distribution found earlier.

[tex]E[X]=\displaystyle\int_0^1xf_X(x)\,\mathrm dx=\frac67\int_0^1(2x^2+x)\,\mathrm dx=\boxed{\frac57}[/tex]

f. This part is cut off, but if you're supposed to find the expectation of [tex]Y[/tex], there are several ways to do so.

Compute the marginal density of [tex]Y[/tex], then directly compute the expected value.

[tex]f_Y(y)=\displaystyle\int_0^1f_{X,Y}(x,y)\,\mathrm dx=\begin{cases}\frac1{14}(4+3y)&\text{for }0<y<2\\0&\text{otherwise}\end{cases}[/tex]

[tex]\implies E[Y]=\displaystyle\int_0^2yf_Y(y)\,\mathrm dy=\frac87[/tex]

Compute the conditional density of [tex]Y[/tex] given [tex]X=x[/tex], then use the law of total expectation.

[tex]f_{Y\mid X}(y\mid x)=\dfrac{f_{X,Y}(x,y)}{f_X(x)}=\begin{cases}\frac{2x+y}{4x+2}&\text{for }0<x<1,0<y,2\\0&\text{otherwise}\end{cases}[/tex]

The law of total expectation says

[tex]E[Y]=E[E[Y\mid X]][/tex]

We have

[tex]E[Y\mid X=x]=\displaystyle\int_0^2yf_{Y\mid X}(y\mid x)\,\mathrm dy=\frac{6x+4}{6x+3}=1+\frac1{6x+3}[/tex]

[tex]\implies E[Y\mid X]=1+\dfrac1{6X+3}[/tex]

This random variable is undefined only when [tex]X=-\frac12[/tex] which is outside the support of [tex]f_X[/tex], so we have

[tex]E[Y]=E\left[1+\dfrac1{6X+3}\right]=\displaystyle\int_0^1\left(1+\frac1{6x+3}\right)f_X(x)\,\mathrm dx=\frac87[/tex]

Final answer:

The question deals with the joint probability density function fX,Y (x, y). It covers verification of the function, calculation of density function of X, probabilities of certain conditions, and finding expectation values of X and Y.

Explanation:

The question revolves around the concept of continuous probability density functions (pdf). Here, the function fX,Y (x, y) is the joint pdf of X and Y under given interval constraints.

(a) To verify if fX,Y (x, y) serves as a joint density function, one would need to confirm that it complies with two conditions: The function should be nonnegative, and the integral over its entire domain should equal to one. You would need to compute a double integral over the range of fX,Y and ascertain if it equals one.

(b)The density function of X can be obtained by integrating fX,Y (x, y) over the range of y i.e., integrate from 0 to 2 for given x.

(c) To compute P(X > Y), the region where this condition holds true needs to be calculated. The marginal density of X must be integrated over this region.

(d) P(Y > 1/2 | X < 1/2) is computed by integrating the joint density function over the region defined by Y > 1/2 and X < 1/2, normalised by the probability of the event X < 1/2.

(e) E(X) is the expectation of X and can be calculated by integrating x*fX(x), where fX(x) is the marginal density function of X.

(f) Similarly, E(Y) can be computed by integrating y*fY(y), with fY(y) being the marginal density of Y.

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Which data set has the highest standard deviation (without doing calculations)?

A) 1,2,3,4

B) 1,1,1,4

C) 1,2,2,4

D) 4,4,4,4

Answers

Answer:

B)1,1,1,4

Step-by-step explanation:

A.1,2,3,4

Mean=[tex]\bar x=\frac{1+2+3+4}{4}=2.5[/tex]

x    [tex]x-\bar x[/tex]   [tex](x-\bar x)^2[/tex]

1       -1.5                                2.25

2       -0.5                              0.25

3         0.5                              0.25  

4          1.5                               2.25

[tex]\sum(x-\bar x)^2=2.25+0.25+0.25+2.25=5[/tex]

B.

Mean=[tex]\bar x=\frac{1+1+1+4}{4}=1.75[/tex]

[tex](x-\bar x)^2[/tex]

0.5625

0.5625

0.5625

5.0625

[tex]\sum(x-\bar x)^2=0.5626+0.5625+0.5625+5.0625=6.75[/tex]

C.

Mean=[tex]\bar x=\frac{1+2+2+4}{4}=2.25[/tex]

[tex](x-\bar x)^2[/tex]

1.5625

0.0625

0.0625

3.0625

[tex]\sum (x-\bar x)=1.5625+0.0625+0.0625+3.0625=4.75[/tex]

D.[tex]Mean=\bar x=\frac{4+4+4+4}{4}=4[/tex]

[tex](x-\bar x)^2[/tex]

0

0

0

0

[tex]\sum (x-\bar x)^2=0+0+0+0=0[/tex]

We know that

S.D is directly proportional to  [tex]\sum (x-\bar x)^2[/tex].

When [tex]\sum (x-\bar x)^2[/tex] is highest  then the S.D is also highest.

We can see that  the value of  [tex]\sum (x-\bar x)^2[/tex] is highest  in option B.

Therefore, S.D of the date set of  option B is highest.

Final answer:

The data set with the numbers 1 through 4 (option A) likely has the highest standard deviation due to having a greater spread compared to others.

Explanation:

The student has asked which data set has the highest standard deviation without doing calculations. In looking at the provided options, we can determine that the higher the variability in the data set, the higher the standard deviation will be. Data set A has numbers 1 through 4, which are more spread out compared to the other sets. Therefore, option A) 1,2,3,4 is likely to have the highest standard deviation. Option D) 4,4,4,4 will have a standard deviation of zero since all the data points are the same. Standard deviation is a measure of how spread out numbers are and a zero standard deviation occurs when all values in a data set are identical.

Assume you have a sample of n1=8, with a sample mean X1= 42 and a sample standard deviation, S1 = 4 and an independent sample of n2=15 from another population with a sample mean of X2=34 and a sample standard deviation of S2 = 34. The pooled variance, Sp 2 = 776.
a. What is the value of the tSTAT for testing H0 : µ1 = µ2?
b. In finding the critical value, how many degrees of freedom are there?
c. What is your statistical decision?

Answers

Answer:

Step-by-step explanation:

Final answer:

To compute the tSTAT, plug the given values into the formula tSTAT = (X1 - X2) / sqrt(Sp2 * (1/n1 + 1/n2)). For the degrees of freedom, calculate df = n1 + n2 - 2. Whether to reject or not the null hypothesis H0 depends on whether the absolute value of tSTAT is greater than the critical value tCritical.

Explanation:

The question pertains to statistical hypothesis testing, specifically the t-test for comparing two independent samples. To solve this:

a. First, calculate the tSTAT as follows: tSTAT = (X1 - X2) / sqrt(Sp2 * (1/n1 + 1/n2)). Using your given values, this becomes tSTAT = (42 - 34) / sqrt(776 * (1/8 + 1/15)). Evaluating the right side will give you the tSTAT.

b. For this t-test, the degree of freedom (df) is calculated as df = n1 + n2 - 2, which gives df = 8+15-2 = 21.

c. The decision whether to reject or fail to reject the null hypothesis H0 depends on the comparison of the calculated tSTAT with the critical value (from the t-distribution table for df = 21). If |tSTAT| > tCritical, reject H0; otherwise, fail to reject H0.

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A billboard designer has decided that a sign advertising the movie Fight Club II - Rage of the Mathematicians should have 1-ft. margins at the top and bottom, and 2-ft. margins on the left and right sides. Furthermore, the billboard should have a total area of 200 ft2 , including the margins. Find the dimensions that would maximize printed area.

Answers

Answer:

Length = 20 ft

Height = 10 ft

Step-by-step explanation:

Let 'X' be the total width of the billboard and 'Y' the total height of the billboard. The total area and printed area (excluding margins) are, respectively:

[tex]200 = x*y\\A_p = (x-4)*(y-2)[/tex]

Replacing the total area equation into the printed area equation, gives as an expression for the printed area as a function of 'X':

[tex]y=\frac{200}{x} \\A_p = (x-4)*(\frac{200}{x} -2)\\A_p=208 -2x -\frac{800}{x}[/tex]

Finding the point at which the derivate for this expression is zero gives us the value of 'x' that maximizes the printed area:

[tex]\frac{dA_p(x)}{dx} =\frac{d(208 -2x -\frac{800}{x})}{dx}=0\\0=-2 +\frac{800}{x^2} \\x=\sqrt{400}\\x=20\ ft[/tex]

If x = 20 ft, then y=200/20. Y= 10 ft.

The dimensions that maximize the printed area are:

Length = 20 ft

Height = 10 ft

Final answer:

The dimensions that maximize the printed area of the billboard are 12 ft by 12 ft (excluding the margins) which gives a printed area of 144 square feet.

Explanation:

To solve this problem, we use the relationship between area, length, and width. The total area, including marginal spaces is given as 200 square feet. The top and bottom margins sum to 2 feet, and the side margins sum to 4 feet.

Let's denote the width of the printed area (excluding the margins) as x and the length as y. Therefore, the width of the whole billboard is x + 4 ft and the length is y + 2 ft.

We know that Area = length * width, so we can establish the equation (x + 4)(y + 2) = 200.

To maximize the printed area, it's best when x and y are equal (a principle in mathematics), making the printed area a square. Also, x + y being constant, the square gives the maximum product.

Solving equation (x + 4) = (y + 2), we find that x = y = 12 feet, which maximizes the printed area giving us 144 square feet.

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A survey randomly selected 250 top executives. The average height of these executives was 66.9 inches with a standard deviation of 6.2 inches. What is a 95% confidence interval for the mean height, μ, of all top executives? a. 63.5 < μ < 66.1 b. 65.3 < μ < 68.5 c. 62.8 < μ < 66.8 d. 66.1 < μ < 67.7

Answers

Answer: Option 'd' is correct.

Step-by-step explanation:

Since we have given that

n = 250

Average = 66.9 inches

standard deviation = 6.2 inches

We need to find the 95% confidence interval for the mean.

So, z = 1.96

Interval would be

[tex]\bar{x}\pm z\dfrac{\sigma}{\sqrt{n}}\\\\=66.9\pm 1.96\times \dfrac{6.2}{\sqrt{250}}\\\\=66.9\pm 0.77\\\\=(66.9-0.77,66.9+0.77)\\\\=(66.13,67.67)[/tex]

Hence, option 'd' is correct.

Researchers claim that 40 tissues is the average number of tissues a person uses during the course of a cold. The company who makes Puffs brand tissues thinks that fewer of their tissues are needed. What are their null and alternative hypotheses?Group of answer choices1. H0: μ < 40 vs. H1: μ = 402. H0: μ = 40 vs. H1: μ < 403. H0: = 40 vs. H1: < 404. H0: μ = 40 vs. H1: μ > 40

Answers

Answer:  2.  H0: μ = 40 vs. H1: μ < 40

Step-by-step explanation:

Null hypothesis [tex](H_0)[/tex] : It is a statement about population parameter by concerning the specific idea. It contains ≤,≥ and = signsAlternative hypothesis [tex](H_a)[/tex] : It is a statement about population parameter but against null hypothesis. It contains < , > and ≠ signs.

Let [tex]\mu[/tex] be the average number of tissues a person uses during the course of a cold.

Given : Researchers claim that 40 tissues is the average number of tissues a person uses during the course of a cold.

i.e. [tex]\mu=40[/tex]

The company who makes Puffs brand tissues thinks that fewer of their tissues are needed.

i.e. [tex]\mu<40[/tex]

Thus , the set of hypothesis would be :-

 [tex]H_0: \mu=40[/tex]

 [tex]H_a: \mu<40[/tex]

Thus , the correct answer is option 2. H0: μ = 40 vs. H1: μ < 40

Final answer:

The null hypothesis (H0) is that an average person uses 40 tissues during a cold (H0: μ = 40). The alternative hypothesis (H1), which Puffs brand wants to prove, is that less than 40 of their tissues are needed during a cold (H1: μ < 40).

Explanation:

The question is related to forming a null hypothesis and an alternative hypothesis in the context of statistics. Here, we are talking about the utilization of tissues during a cold. The researchers claim that on average a person uses 40 tissues (let's call this number 'μ') throughout a cold's duration. However, the Puffs brand makers, as per their assumption, think that less tissues of their brand are needed.

The null hypothesis, denoted by H0, is usually a claim of no effect or no difference. In this case, it is H0: μ = 40, denoting that the average number of tissues used is 40. This is also the claim that the Puffs brand intends to challenge or nullify.

On the other hand, the alternative hypothesis (H1) is the claim that the party interested in the outcome (here, the Puffs brand) wants to validate. Here, it would be H1: μ < 40, which signifies that fewer than 40 Puffs tissues are needed during the course of a cold.

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Farmer Pickles wants Bob to paint the circular fence which encloses his sunflower field. If the parametric equations x = 18 cos(θ) and y = 18 sin(θ) describe the base of the fence (in yards) and the height of the fence is given by the equation h(x, y) = 12 + (2x − y)/6, then how many gallons of paint will Bob need to complete the project. Assume that one gallon of paint covers three hundred square feet of fence.

Answers

Answer:

40.72 gallons

Step-by-step explanation:

Since [tex]x = 18cos(\theta)[/tex] and [tex]y = 18sin(\theta)[/tex]. We can say this base has a shape of a circle of radius r = 18.

We can also calculate h in term of angle θ:

[tex]h(\theta) = 12 + \frac{36cos(\theta) - 18sin(\theta)}{6}[/tex]

[tex]h(\theta) = 12 + 6cos(\theta) - 3sin(\theta)[/tex]

We can calculate the area of the fence if we integrate this h function over θr, which is the chord length. θ ranges from 0 to 2π

A = ∫h(θ)rdθ

A = ∫(12 + 6cos(θ) - 3sin(θ))18d

A = (12θ + 6sin(θ) + 3cos(θ))18

As θ ranges from 0 to 2π as a full circle

A = (12*0 + 6sin(0) + 3cos(0))18 - (12*2π + 6sin(2π) + 3cos(2π))18

A = 3*18 - 18(24π + 3) = 1357.17 square yard

As 1 yard = 3 feet then 1 square yard = 9 square feet

1 gallon of paint can cover 300 square feet or 300 / 9 = 33.33 square yard

So to cover square yard Bob would need

1357.17 / 33.33 = 40.72 gallons of paint.

A baseball player bunts a ball down the first base line. It rolls 33 ft at an angle of 25 Degree with the first base path. The pitcher's mound is 60.5 ft from home plate. How far must he travel to get to the ball? Note that a baseball diamond is a square. The pitcher must run feet. (Round to the nearest foot.)

Answers

Final answer:

The distance the pitcher must travel to get to the ball is approximately 36 ft.

Explanation:

The distance the pitcher must travel to get to the ball can be found using trigonometry. We can break down the motion of the ball into horizontal and vertical components. The horizontal component of the motion is the distance the ball rolls down the first base line, which is given as 33 ft. The vertical component of the motion can be found using the angle of 25 degrees. We can use the sine function to find the vertical distance:

Vertical distance = sin(25 degrees) * 33 ft = 14.04 ft

The distance the pitcher must travel is the hypotenuse of a right triangle formed by the horizontal and vertical distances. Using the Pythagorean theorem, we can find the distance:

Distance = sqrt((33 ft)^2 + (14.04 ft)^2) = 35.88 ft

Therefore, the pitcher must run approximately 36 ft to get to the ball.

Hypothetical. A box is full of thousands of tickets labeled either O or 1 It is believed that the ave age of all he tickets in the box that ·the ro tion of S S 020 or 2 To test the null hypothesis that 20% of the tickets are labeled 1, we draw 400 tickets at random. Of these, 102 are labeled 1. Round your values to three decimal places. You can choose to work with the sum or the average (that is, the number of 1's or the proportion of 1's); the test statistic will be the same. Use this problem to practice this idea. If the null hypothesis is right, then the expected value of the number of 1's in 400 random draws from this box is If the null hypothesis is right, then the expected value of the proportion of 1's in 400 random draws from this box is with a standard error of The observed number of 1's is The observed proportion of 1's is 0.255 with a standard error of The test statistic is 2.75 The P-value (using a 2-test) is approximately 0.006

Answers

Answer: 28 * 1

Step-by-step explanation:

The height and radius of a cone are each multiplied by 3. What effect does this have
on the volume of the cone?
The volume of the cone is multiplied by..​

Answers

The volume of the original cone is multiplied by 27

Solution:

We have been given that the height and radius of a cone are each multiplied by 3.

We need to find what effect it has on the volume of the cone.

The volume of the original cone is given as follows:

[tex]\text {Volume of cone}=\frac{1}{3} \pi r^{2} h[/tex]

Now, the radius and height are multiplied by 3.

Therefore, the new radius is ‘3r’ and the new height is ‘3h’.

Hence the new volume is  

[tex]\text {Volume of new cone}=\frac{1}{3} \pi(3 r)^{2}(3 h)=9 \pi r^{2} h[/tex]

original volume of cone x 27 = volume of new cone

Therefore, the original volume of cone is multiplied by 27

Final answer:

When the height and radius of a cone are each multiplied by 3, the volume of the cone is multiplied by 27.

Explanation:

When the height and radius of a cone are each multiplied by 3, the effect on the volume of the cone is that the volume is multiplied by 27.

The formula for the volume of a cone is V = (1/3)πr²h. When the height and radius are multiplied by 3, the new volume would be V' = (1/3)π(3r)²(3h) = 27(1/3)πr²h = 27V.

Therefore, the volume of the cone is multiplied by 27.

The velocity function (in meters per second) is given for a particle moving along a line.v(t) = 3t − 8, 0 ≤ t ≤ 3(a) Find the displacement.(b) Find the distance traveled by the particle during the given time interval.

Answers

(a) The displacement of the particle is [tex]\(-\frac{21}{2}\)[/tex] meters.

(b) The distance travelled by the particle during the given time interval is [tex]\(\frac{169}{6}\)[/tex] meters.

(a) To find the displacement of the particle over the time interval [0, 3], we need to calculate the definite integral of the velocity function v(t) over this interval. The displacement is given by the integral of velocity with respect to time.

The velocity function is v(t) = 3t - 8.

The displacement S over the interval [a, b] is given by:

[tex]\[ S = \int_{a}^{b} v(t) \, dt \][/tex]

For our interval [0, 3], we have:

[tex]\[ S = \int_{0}^{3} (3t - 8) \, dt \][/tex]

To compute this integral, we find the antiderivative of the integrand:

[tex]\[ \int (3t - 8) \, dt = \frac{3}{2}t^2 - 8t + C \][/tex]

Now, we evaluate this antiderivative at the endpoints of the interval and subtract:

[tex]\[ S = \left(\frac{3}{2}(3)^2 - 8(3)\right) - \left(\frac{3}{2}(0)^2 - 8(0)\right) \][/tex]

[tex]\[ S = \left(\frac{3}{2} \cdot 9 - 24\right) - (0) \][/tex]

[tex]\[ S = \frac{27}{2} - 24 \][/tex]

[tex]\[ S = \frac{27}{2} - \frac{48}{2} \][/tex]

[tex]\[ S = -\frac{21}{2} \][/tex]

So the displacement of the particle is [tex]\(-\frac{21}{2}\)[/tex] meters.

(b) To find the distance travelled by the particle, we need to consider the intervals where the velocity is positive and where it is negative. The distance travelled is the sum of the distances travelled during each interval, without regard to direction.

First, we find the times when the velocity is zero by setting v(t) equal to zero:

[tex]\[ 3t - 8 = 0 \][/tex]

[tex]\[ 3t = 8 \][/tex]

[tex]\[ t = \frac{8}{3} \][/tex]

Since [tex]\(0 \leq t \leq 3\),[/tex] the velocity changes sign at [tex]\(t = \frac{8}{3}\)[/tex]. This divides the interval [0, 3] into two subintervals: [0, 8/3] and [8/3, 3].

We calculate the distance travelled in each subinterval and sum them:

For [0, 8/3]:

[tex]\[ d_1 = \int_{0}^{\frac{8}{3}} |3t - 8| \, dt \][/tex]

Since the velocity is negative on this interval, we have:

[tex]\[ d_1 = \int_{0}^{\frac{8}{3}} (8 - 3t) \, dt \][/tex]

[tex]\[ d_1 = \left(8t - \frac{3}{2}t^2\right) \Bigg|_{0}^{\frac{8}{3}} \][/tex]

[tex]\[ d_1 = \left(8\left(\frac{8}{3}\right) - \frac{3}{2}\left(\frac{8}{3}\right)^2\right) - (0) \][/tex]

[tex]\[ d_1 = \frac{64}{3} - \frac{3}{2}\left(\frac{64}{9}\right) \][/tex]

[tex]\[ d_1 = \frac{64}{3} - \frac{32}{3} \][/tex]

[tex]\[ d_1 = \frac{32}{3} \][/tex]

For [8/3, 3]:

[tex]\[ d_2 = \int_{\frac{8}{3}}^{3} |3t - 8| \, dt \][/tex]

Since the velocity is positive on this interval, we have:

[tex]\[ d_2 = \int_{\frac{8}{3}}^{3} (3t - 8) \, dt \][/tex]

[tex]\[ d_2 = \left(\frac{3}{2}t^2 - 8t\right) \Bigg|_{\frac{8}{3}}^{3} \][/tex]

[tex]\[ d_2 = \left(\frac{3}{2}(3)^2 - 8(3)\right) - \left(\frac{3}{2}\left(\frac{8}{3}\right)^2 - 8\left(\frac{8}{3}\right)\right) \][/tex]

[tex]\[ d_2 = \left(\frac{27}{2} - 24\right) - \left(\frac{32}{3} - \frac{64}{3}\right) \][/tex]

[tex]\[ d_2 = \frac{27}{2} - 24 + \frac{32}{3} \][/tex]

[tex]\[ d_2 = \frac{81}{6} - \frac{144}{6} + \frac{64}{6} \][/tex]

[tex]\[ d_2 = -\frac{105}{6} \][/tex]

[tex]\[ d_2 = -\frac{35}{2} \][/tex]

Since we cannot have a negative distance, we take the absolute value:

[tex]\[ d_2 = \frac{35}{2} \][/tex]

The total distance travelled is the sum of the distances in each interval:

[tex]\[ D = d_1 + d_2 \][/tex]

[tex]\[ D = \frac{32}{3} + \frac{35}{2} \][/tex]

[tex]\[ D = \frac{64}{6} + \frac{105}{6} \][/tex]

[tex]\[ D = \frac{169}{6} \][/tex]

So the distance travelled by the particle during the given time interval is [tex]\(\frac{169}{6}\)[/tex] meters.

How do I solve this?

Answers

Answer:

[tex]t=\frac{\pm\sqrt{dk}}{k}[/tex]

Step-by-step explanation:

Given relation:

[tex]d=k\ t^2[/tex]

We need to solve for [tex]t[/tex]

The given relation can be rearranged by isolating [tex]t[/tex] on one side.

Swapping the sides of the equation we have,

[tex]k\ t^2=d[/tex]

Dividing both sides by [tex]k[/tex] on order to cancel out [tex]k[/tex] on left side.

[tex]\frac{k\ t^2}{k}=\frac{d}{k}[/tex]

[tex]t^2=\frac{d}{k}[/tex]

We have got an expression for [tex]t^2[/tex] but we need to solve for [tex]t[/tex]

So, we take square root both sides to change [tex]t^2[/tex] to [tex]t[/tex]

[tex]\sqrt{t^2}=\sqrt{\frac{d}{k}}[/tex]

[tex]t=\pm\sqrt{\frac{d}{k}}[/tex]

So, we have successfully isolated [tex]t[/tex] on left side.

But the expression we got is a fraction with a square root  in the denominator. Thus we need to rationalize it to make it in simplest form.

The expression can be written by taking square root separately for numerator and denominator :

[tex]t=\pm{\frac{\sqrt{d}}{\sqrt{k}}}[/tex]

Multiplying the numerator and denominator by [tex]\sqrt{k}[/tex]

[tex]t=\pm\frac{\sqrt{d}}{\sqrt{k}}\times\frac{\sqrt{k}}{\sqrt{k}} [/tex]

[tex]t=\pm\frac{\sqrt{dk}}{(\sqrt{k})^2}[/tex]

Square of a square root will remove the square root. Thus we have,

∴ [tex]t=\frac{\pm\sqrt{dk}}{k}[/tex]

Thus we have successfully got the expression in the simplest form.

Two points are selected randomly on a line of length 40 so as to be on opposite sides of the midpoint of the line. In other words, the two points X and Y are independent random variables such that X is uniformly distributed over [0,20) and Y is uniformly distributed over(20,40]. Find the probability that the distance between the two points is greater than 12 .

Answers

Answer:

Probability that we have an ordered pair (x,y) representing two points that satisfy the conditions is 328/400.

Step-by-step explanation:

This is a geometric probability.

set of all possible x in [0,20) and y in (20,40] we obtain a square in the x-y plane (in the first quadrant).

Area of this square is 20x20 = 400.

Set of all (x,y), that satisfy the distance being greater than 12

y > x, this means that (y - x) > 12 which is the same as the inequality y > x + 12.

For inequality, we obtain the region in the plane above the line y = x + 12.  The area of this region is

(1/2)bh = (1/2)(12)(12) = 72.

Thus the area of the set of all points in our square that satisfy the condition (y - x) > 12 is 400 - 72 = 328.

 Probability that we have an ordered pair (x,y) representing two points that satisfy the conditions is 328/400.

What is the result when the number 30 is decreased by 10%

Answers

Final answer:

The result when the number 30 is decreased by 10% is 27, calculated by finding 10% of 30, which is 3, and subtracting it from the original number.

Explanation:

When the number 30 is decreased by 10%, this implies we need to find 10% of 30 and then subtract it from 30. To calculate 10% of a number, you can simply divide the number by 10. Hence, 10% of 30 is 30 ÷ 10 = 3. Now, to decrease 30 by this value, we do the subtraction: 30 - 3 = 27. Therefore, the result when the number 30 is decreased by 10% is 27.

The number 27 is obtained by subtracting 10% from 30.

Step 1: Identify the initial number.

Given that the initial number is 30.

Step 2: Determine the percentage to be decreased.

The percentage to be decreased is 10%.

Step 3: Calculate the amount to be decreased.

To find 10% of 30, multiply 30 by 0.10.

10% of 30 = 30 × 0.10 = 3.

Step 4: Subtract the amount to be decreased from the initial number.

30 - 3 = 27.

Step 5: Interpret the result.

So, when 30 is decreased by 10%, the result is 27.

Therefore, the result of decreasing the number 30 by 10% is 27.

A precision control engineer is interested in the mean length of tubing being cut automatically by machine. It is known that the standard deviation in the cutting length is 0.15 feet. A sample of 60 cut tubes yields a mean length of 12.15 feet. This sample will be used to obtain a 99% confidence interval for the mean length cut by machine. Develop the 99% confidence interval for population mean.

Answers

Answer:

The 99% confidence interval would be given by (12.10;12.20)

Step-by-step explanation:

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

[tex]\bar X=12.15[/tex] represent the sample mean  

[tex]\mu[/tex] population mean (variable of interest)

[tex]\sigma=0.15[/tex] represent the population standard deviation

n=60 represent the sample size  

Assuming the X follows a normal distribution

[tex]X \sim N(\mu, \sigma=0.15[/tex]

The sample mean [tex]\bar X[/tex] is distributed on this way:

[tex]\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})[/tex]

The confidence interval on this case is given by:

[tex]\bar X \pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}[/tex]   (1)

The next step would be find the value of [tex]\z_{\alpha/2}[/tex], [tex]\alpha=1-0.99=0.01[/tex] and [tex]\alpha/2=0.005[/tex]

Using the normal standard table, excel or a calculator we see that:

[tex]z_{\alpha/2}=2.58[/tex]

Since we have all the values we can replace:

[tex]12.15 - 2.58\frac{0.15}{\sqrt{60}}=12.10[/tex]  

[tex]12.15 + 2.58\frac{0.15}{\sqrt{60}}=12.20[/tex]  

So on this case the 99% confidence interval would be given by (12.10;12.20)  

A password can be any string of length 7, 8, or 9. Each character in the password can be any capital letter or special character from the set {*, $, &, #, %}. There are no other restrictions on the password. How many possible different passwords are there?

Answers

Answer:

744

Step-by-step explanation:

The alphabet has 26 letters. Since each character of the password can be a capital letter or one of the four character of the set {*, $, &, #, %}, then each space in the password has 26+5=31 options to select a character.

a) If the password has length 7, then there are 31*7=217 different passwords.

b) If the password has length 8, then there are 31*8=248 different passwords and,

c) if the password has length 9, then there are 31*9=279 different passwords.

Then the total of different passwords is 217+248+279=744.

Final answer:

There are different numbers of possible passwords based on the length of the password.

Explanation:

To determine the number of possible different passwords, we need to consider the different options for each character in the password and multiply them together.

For a password of length 7, each character has 5 possible options (capital letters or special characters). Therefore, there are 5 options for each of the 7 characters, giving us a total of 5^7 = 78,125 possible passwords.

Similarly, for a password of length 8, there are 5 options for each of the 8 characters, giving us a total of 5^8 = 390,625 possible passwords.

Finally, for a password of length 9, there are 5 options for each of the 9 characters, giving us a total of 5^9 = 1,953,125 possible passwords.

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Suppose the interval ​[negative 2​,0​] is partitioned into nequals4 subintervals. What is the subinterval length Upper Delta x​? List the grid points x0​, x1​, x2​, x3​, x4. Which points are used for the​ left, right, and midpoint Riemann​ sums?

Answers

Answer:

We find the length of each subinterval dividing the distance between the endpoints of the interval by the quantity of subintervals that we want.

Then

Δx= [tex]\frac{0-(-2)}{4}=\frac{2}{4}=\frac{1}{2}[/tex]

Now, each [tex]x_i[/tex] is found by adding Δx iteratively from the left end of the interval.

So

[tex]x_0=-2\\x_1=-2+\frac{1}{2}=\frac{-3}{2}\\x_2=\frac{-3}{2}+\frac{1}{2}=-1\\x_3=-1+\frac{1}{2}=-\frac{1}{2}\\x_4=\frac{-1}{2}+\frac{1}{2}=0[/tex]

Each subinterval is

[tex]s_1=[-2,-3/2]\\s_2=[-3/2,-1]\\s_3=[-1,-1/2]\\s_4=[-1/2,0][/tex]

The midpoints of the subintervals are

[tex]m_1=\frac{-2-3/2}{2}=\frac{-7/2}{2}=\frac{-7}{4}\\m_2=\frac{-1-3/2}{2}=\frac{-5/2}{2}=\frac{-5}{4}\\m_3=\frac{-1/2-1}{2}=\frac{-3/2}{2}=\frac{-3}{4}\\m_4=\frac{0-1/2}{2}=\frac{-1}{4}[/tex]

The points used for the

1. left Riemann sums are the left endpoints of the subintervals, that is

[tex]x_0=-2, x_1=\frac{-3}{2}, x_2=-1, x_3= \frac{-1}{2}[/tex]

2. right Riemann sums are the right endpoints of the subinterval,

[tex]x_1=-\frac{3}{2}, x_2=-1, x_3=-\frac{1}{2}, x_4=0[/tex]

3. midpoint Riemann sums are the midpoints of each subinterval

[tex]m_1,m_2,m_3,m_4.[/tex]

Final answer:

The subinterval length in the given range is 0.5. The grid points are -2, -1.5, -1, -0.5, and 0. The points used for the left, right, and midpoint Riemann sums vary based on the position of the subinterval.

Explanation:

The interval specified is ​[negative 2​,0​] and is partitioned into n=4 subintervals. The length of each subinterval, Upper Delta x, is determined by subtracting the lower boundary from the upper boundary and dividing by the number of subintervals, in this case (0 - (-2)) / 4 = 0.5. Therefore, the subinterval length is 0.5.

The grid points are calculated by adding the subinterval length to each previous point, starting from the lower boundary: x0​ is -2, for x1, add the subinterval length 0.5 to x0, resulting in -1.5. Repeat this process to find x2, x3 and x4 which are -1, -0.5 and 0 respectively.

Now, for the left Riemann sum, you use the left-hand side points of each subinterval, which are x0​, x1​, x2​, x3. The right Riemann sum uses the right-hand side points, x1​, x2​, x3​, x4. Lastly, the midpoint Riemann sum uses points in the middle of each subinterval. As the subintervals here each have a length of 0.5, their midpoints fall on x0.5, x1.5, x2.5 and x3.5 respectively.

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Two vertical poles, one 4 ft high and the other 16 ft high, stand 15 feet apart on a flat field.
A worker wants to support both poles by running rope from the ground to the top of each post.

If the worker wants to stake both ropes in the ground at the same point, where should the stake be placed to use the least amount of rope?

Answers

Answer:

The rope should be staked at 3ft  distance from bottom of  4 ft pole.

Step-by-step explanation:

Given that  the poles are 4 ft and 16 ft, separated by distance of 15 ft.

Refering the given figure, let the distance from 4 ft pole be "x".

consequently, distance from 16 ft pole is (15-x).

Refering the figure, in right angled triangle DCO, by pythagoras theorm, the the length OD is [tex]\sqrt{16-x^{2} }[/tex].

Similarly, the length OA is [tex]\sqrt{481+x^{2}-30x}[/tex].

thus the length of rope is AO + OD =  L = f(x) =

[tex]\sqrt{16-x^{2} }[/tex] + [tex]\sqrt{481+x^{2}-30x}[/tex]

now, for a function of one variable, to have maximum or minimum, differentiate once, and find value of x.

Now, f'(x)= [tex]x(\sqrt{16+x^{2} }^{-1} )+ (x-15)(\sqrt{481+x^{2}-30x }^{-1})[/tex]

f'(x)=0  gives,

x=3 or -5

but the length cannot be negetive.

thus x=3.

By applying the pythagroean theorem and differentiation, the distance from the bottom of the pole that the rope should be staked at is: 3 ft.

What is the Pythagorean Theorem?

The pythagorean theorem states that the square of the hypotenuse of a right triangle equals the sum the square of the otehr two legs of the right triangle.

The sketch of the problem is given in the diagram that is attached below.

Applying the pythagorean theorem in right triangle DCO, we would have:

DO = √(16 - x²)

AO would also be found using the pythagorean theorem:

AO = √(481 + x² - 30x)

Therefore, the length of the rope = AO + DO = f(x), which is:

f(x) = √(481 + x² - 30x) + √(16 - x²)

To find the value of x, take derivative and set equal to 0.

Differentiating, f(x) = √(481 + x² - 30x) + √(16 - x²), we would have:

f'(x) = 0 which will give us, x = 3 or -5.

Since the length cannot be negative, therefore, the distance from the bottom of the pole that the rope should be staked at is: 3 ft.

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Factor the polynomial completely. 8x^4y – 16x^2y^2

Answers

8x^4y – 16x^2y^2=

=8x^4y-16x^4y

=-8x^4y

=-8*(x^4y)

Answer:

8\,x^2\,y\,(x-\sqrt{2} )(x+\sqrt{2} )

Step-by-step explanation:

Let's start by extracting all common factors from the two terms of this binomial. These common factors are: 8, [tex]x^2[/tex], and [tex]y[/tex].

The extraction renders:

[tex]8\,x^2\,y\,(x^2-2)[/tex]

In the real number system, the binomial in parenthesis can still be factored out considering that 2 is the perfect square of [tex]\sqrt{2}[/tex], that is:

[tex]2=(\sqrt{2} )^2[/tex]

We can then forwards with the factoring of this binomial using the factorization of a difference of squares as:

[tex](x^2-2) = (x^2-(\sqrt{2} )^2)=(x-\sqrt{2} )(x+\sqrt{2} )[/tex]

Thus giving the complete factorization as:

[tex]8\,x^2\,y\,(x-\sqrt{2} )(x+\sqrt{2} )[/tex]

A report describes the results of a large survey involving approximately 3500 people that was conducted for the Center for Disease Control. The sample was selected in a way that the Center for Disease Control believed would result in a sample that was representative of adult Americans. One question on the survey asked respondents if they had learned something new about a health issue or disease from a TV show in the previous 6 months. Data from the survey was used to estimate the following probabilities, where L-event that a randomly selected adult American reports learning somethlng new about a health Issue or disease from a TV show in the previous 6 months and F-event that a randomly selected adult American is female Assume that P(F) 0.5. Are the events L and F independent events? Use probabilities to justify your answer. Land Flare notindependent events, because P(L) P(F)- 0.3 , which is not equal to P(L n F)

Answers

0.3 - 60 is your right answer!

A certain vibrating system satisfies the equation . Find the value of the damping coefficient for which the quasi period of the damped motion is greater than the period of the corresponding undamped motion. Round you answer to three decimal places.

Answers

Answer:

Hence the value of damping coefficient = 1.49071.

Step-by-step explanation:

Some manufacturers claim that non-hybrid sedan cars have a lower mean miles-per-gallon (mpg) than hybrid ones. Suppose that consumers test 21 hybrid sedans and get a mean of 32 mpg with a standard deviation of 8 mpg. Thirty-one non-hybrid sedans get a mean of 21 mpg with a standard deviation of three mpg. Suppose that the population standard deviations are known to be six and three, respectively. Conduct a hypothesis test at the 5% level to evaluate the manufacturers claim.

Answers

I got to think about this again.. Come back later!! X=22.23

A polling organization took a random sample of 2,500 single parents to determine how many single parents have two jobs. Suppose 38% of all single parents have two jobs. If the total population of single parents is 12 million, is the 10% condition met? Justify your answer.

Answers

Answer:

Yes it is below 10% of the population

Step-by-step explanation:

10% of 12 million is 1,200,00

One of the conditions is as 2,500

The population is at least 10 times as large as the sample.

here 10*n = 10* 2500 = 25000 < 12000000

So the correct option is "Yes, 10(2,500) = 25,000, which is less than the total population."

5. The Lakeland Post polled 1,231 adults in the city to determine whether they wash their hair before their body while in the shower. Of the respondents, 63% said they washed their hair first. Suppose 51% of all adults actually wash their hair first. What are the mean and standard deviation of the sampling distribution?

Sampling distribution of sample proportion ( \hat p ) is approximately normal for large n with mean and standard deviations are as

mean = p = 0.51

for finding the standard deviation we need to use p = 0.51

б = √p·(1 - p )/ n = √0.51 . 0.49/ 1234 = 0.014248

So the correct choice is "The mean is 0.51 and the standard deviation is 0.0142"

Learn more about polling organizations at

https://brainly.com/question/971724

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One of two small classrooms is chosen at random with equally likely probability, and then a student is chosen at random from the chosen classroom.
Classroom #1 has 5 boys and 13 girls.
Classroom #2 has 14 boys and 9 girls.

What is the probability that Classroom #2 was chosen at random, given that a girl was chosen?

Answers

Answer: Our required probability is 0.351.

Step-by-step explanation:

Since we have given that

Probability of getting classroom 1 = [tex]\dfrac{1}{2}[/tex]

Probability of getting classroom 2 = [tex]\dfrac{1}{2}[/tex]

Probability of getting girl from classroom 1 = [tex]\dfrac{13}{18}[/tex]

Probability of getting girl from classroom 2 = [tex]\dfrac{9}{23}[/tex]

Using "Bayes theorem".

so, the probability that classroom 2 was chosen given that a girl was chosen would be

[tex]\dfrac{\dfrac{1}{2}\times \dfrac{9}{23}}{\dfrac{1}{2}\times \dfrac{9}{23}+\dfrac{1}{2}\times \dfrac{13}{18}}\\\\=\dfrac{0.195}{0.195+0.361}\\\\=0.351[/tex]

Hence, our required probability is 0.351.

There are five Oklahoma State Officials: Governor (G), Lieutenant Governer (L), Secretary of State (S), Attorney General (A), and Treasurer (T). Take all possible samples without replacement of size 3 that can be obtained from the population of five officials. (Note, there are 10 possible samples!)
(a) What is the probability that the governor is included in the sample?
(b) What is the probability that the governor and the attorney general are included in the sample?

Answers

So, the required probabilities are,

Part(a):P(A)=0.6

Part(b):P(B)=0.1

Given that:

There are five Oklahoma state officials,

Governor (G), Lieutenant Governor (L), Secretary of State (S), Attorney General (A), and Treasurer (T).

All possible samples of size 3 are obtained from the population of five officials.

Here order does not matter so we use the combinations.

[tex]5_C_3=10[/tex] possible samples.

So, S={GLS,GLA,GLT,GSA,GST,GAT,LSA,LST,LAT,SAT}

Hence, n(s)=10

Part(a):

Let A denotes the event that the governor is included in the sample.

A={GLS,GLA,GLT,GSA,GST,GAT}

That is n(A)=6

So, the probability that the governor is included in the sample,

[tex]P(A)=\frac{n(A)}{n(S)} \\=\frac{6}{10}\\ =0.6[/tex]

Part(b):

Let B denotes the event that the government attorney general and the treasure are included in the sample.

B={GAT}

That is n(B)=1

Hence, the probability that the government attorney general and the treasure are included in the sample is,

[tex]P(B)=\frac{n(B)}{n(S)} \\=\frac{1}{10} \\=0.1[/tex]

Learn More:https://brainly.com/question/6077878

(a) Probability of governor included: 6/10 = 0.6.

(b) Probability of governor and attorney general included: 1/10 = 0.1.

let's break it down in detail:

(a) Probability that the governor is included in the sample:

To find this probability, we need to count how many of the possible samples include the governor. From the list of all possible samples:

1. {G, L, S}

2. {G, L, A}

3. {G, L, T}

4. {G, S, A}

5. {G, S, T}

6. {G, A, T}

We see that in 6 out of 10 samples, the governor is included. Thus, the probability of the governor being included in the sample is [tex]\( \frac{6}{10} = 0.6 \).[/tex]

(b) Probability that the governor and the attorney general are included in the sample:

To find this probability, we need to count how many of the possible samples include both the governor and the attorney general. Looking at the list of all possible samples again:

1. {G, L, S}

2. {G, L, A}

3. {G, L, T}

4. {G, S, A}

5. {G, S, T}

6. {G, A, T}

7. {L, S, A}

8. {L, S, T}

9. {L, A, T}

10. {S, A, T}

We see that only in one out of the 10 samples, both the governor and the attorney general are included (sample number 2). Thus, the probability of both the governor and the attorney general being included in the sample is [tex]\( \frac{1}{10} = 0.1 \).[/tex]

So, to summarize:

(a) Probability that the governor is included in the sample: 0.6

(b) Probability that the governor and the attorney general are included in the sample: 0.1

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