The recommended daily dietary allowance for zinc among males older than age 50 years is 15 mg/day. An article reports the following summary data on intake for a sample of males age 65−74 years: n = 114, x = 11.2, and s = 6.58. Does this data indicate that average daily zinc intake in the population of all males age 65−74 falls below the recommended allowance? (Use α = 0.05.) State the appropriate null and alternative hypotheses.

Answer:

-1/3

Step-by-step explanation:

rise / run = slope

4-5/ 1-(-2)

= -1/3

Answer: she is 8.06 miles from her starting point

Step-by-step explanation:

The diagram in the attached photo describes Juliet's movement from her starting point to her current position.

A triangle ABC is formed

AB = distance that she walked towards west

BC = distance that she walked towards south

AC= x = distance that she is from her starting point

The diagram is a right angle triangle. So we can find the distance that she is from her starting point using Pythagoras theorem.

Hypotenuse^2 = opposite^2 + adjacent^2

Hypotenuse = x

Opposite = 7

Adjacent = 4

x^2 = 7^2 + 4^2

x^2 = 49 + 16 = 65

x = √65 = 8.06 miles

The question is about calculating the mean and standard deviation of a discrete random variable representing the size of the freezer model a customer buys, and then the mean and standard deviation of the price paid, which is a function of the freezer size.

The random variable X in this scenario represents the size of the freezer model (in cubic feet) that the next customer buys from the appliance dealer. This is a discrete random variable, with possible values being the sizes of the three available freezer models: 14.5, 16.9, and 19.1 cubic feet.

To calculate the mean and standard deviation of X, we can use the formula for the mean of a discrete random variable and the formula for the standard deviation respectively, which are given as follows:
Mean (expected value), E(X) = Σ [x * p(x)]
Standard Deviation, σ(X) = sqrt{Σ [(x - μ)² * p(x)]}

The mean value of the second variable (Price) can be calculated by plugging the expected value of X into the given Price = 25X - 8.5 equation, and the standard deviation by applying the transformation rule for standard deviations (σ_Y = |b| * σ_X, where b is the coefficient of X in the original expression for Y, which in this case is 25).

https://brainly.com/question/17238412

#SPJ2

Answer:

\mu = 14.5\\

\sigma = 5.071\\

k = 1.084

Step-by-step explanation:

given that a  statistician uses Chebyshev's Theorem to estimate that at least 15 % of a population lies between the values 9 and 20.

i.e. his findings with respect to probability are

[tex]P(9<x<20) \geq 0.15\\P(|x-14.5|<5.5) \geq 0.15[/tex]

Recall Chebyshev's inequality that

[tex]P(|X-\mu |\geq k\sigma )\leq {\frac {1}{k^{2}}}\\P(|X-\mu |\leq k\sigma )\geq 1-{\frac {1}{k^{2}}}\\[/tex]

Comparing with the Ii equation which is appropriate here we find that

[tex]\mu =14.5[/tex]

Next what we find is

[tex]k\sigma = 5.5\\1-\frac{1}{k^2} =0.15\\\frac{1}{k^2}=0.85\\k=1.084\\1.084 (\sigma) = 5.5\\\sigma = 5.071[/tex]

Thus from the given information we find that

[tex]\mu = 14.5\\\sigma = 5.071\\k = 1.084[/tex]

Answer: 90% confidence interval would be (6.022,6.986).

Step-by-step explanation:

Since we have given that

Number of students = 361

Sample mean = 6.504

Sample standard deviation = 5.584

Standard error of the mean = 0.294

At 90% confidence level,

So, α = 0.01

So, z = 1.64

Margin of error is given by

[tex]z\times \text{Standard error}\\\\=1.64\times 0.294\\\\=0.48216[/tex]

So, Lower limit would be

[tex]\bar{x}-0.482\\\\=6.504-0.482\\\\=6.022[/tex]

Upper limit would be

[tex]\bar{x}+0.482\\\\=6.504+0.482\\\\=6.986[/tex]

So, 90% confidence interval would be (6.022,6.986).

Answer: Yes, this suggest that the actual percentage of type A donations differs from 40%, the percentage of the population having type A blood.

Step-by-step explanation:

Since we have given n = 157

x = 86

So, [tex]\hat{p}=\dfrac{x}{n}=\dfrac{86}{157}=0.55[/tex]

and we have p = 0.4

So, hypothesis would be

[tex]H_0:p=\hat{p}\\\\H_a:p\neq \hat{p}[/tex]

Since there is 1% level of significance.

So, test statistic value would be

[tex]z=\dfrac{\hat{p}-p}{\sqrt{\dfrac{p(1-p)}{n}}}\\\\z=\dfrac{0.55-0.40}{\sqrt{\dfrac{0.4\times 0.6}{157}}}\\\\z=\dfrac{0.15}{0.039}\\\\z=3.846[/tex]

and the critical value at 1% level of significance , z = 2.58

Since 2.58<3.846.

So, we reject the null hypothesis.

Hence, Yes, this suggest that the actual percentage of type A donations differs from 40%, the percentage of the population having type A blood.

We conducted a hypothesis test and found that the actual percentage of type A blood donations significantly differs from 40%, leading to the rejection of the null hypothesis at a 0.01 significance level.

Hypothesis Testing for Blood Type Proportion

We will perform a hypothesis test to determine whether the percentage of type A blood donations differs from 40%. The appropriate hypotheses for this test are:

Next, we calculate the test statistic for the sample proportion:

The critical value for a two-tailed test at a significance level of 0.01 is approximately ±2.576.

Since 3.79 > 2.576, we reject the null hypothesis (H0). Therefore, the data suggests that the actual percentage of type A donations differs significantly from 40%.

Answer:

The p value obtained was a low value and using the significance level given [tex]\alpha=0.05[/tex] we have [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the proportion of of driver fatalities related to alcohol is less from 0.77 or 77%.  

Step-by-step explanation:

1) Data given and notation n  

n=53 represent the random sample taken

X=35 represent the automobile driver fatalities in a certain county involved with an intoxicated driver

[tex]\hat p=\frac{35}{53}=0.660[/tex] estimated proportion of automobile driver fatalities in a certain county involved with an intoxicated driver

[tex]p_o=0.77[/tex] is the value that we want to test

[tex]\alpha=0.05[/tex] represent the significance level

Confidence=95% or 0.95

z would represent the statistic (variable of interest)

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

2) Concepts and formulas to use  

We need to conduct a hypothesis in order to test the claim that the population proportion of driver fatalities related to alcohol is less than 77% or 0.77 in Kit Carson:  

Null hypothesis:[tex]p\geq 0.77[/tex]  

Alternative hypothesis:[tex]p < 0.77[/tex]  

When we conduct a proportion test we need to use the z statisitc, and the is given by:  

[tex]z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}[/tex] (1)  

The One-Sample Proportion Test is used to assess whether a population proportion [tex]\hat p[/tex] is significantly different from a hypothesized value [tex]p_o[/tex].

3) Calculate the statistic  

Since we have all the info requires we can replace in formula (1) like this:  

[tex]z=\frac{0.660 -0.77}{\sqrt{\frac{0.77(1-0.77)}{53}}}=-1.903[/tex]

4) Statistical decision  

It's important to refresh the p value method or p value approach . "This methos is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.  

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

Since is an unilateral lower test the p value would be:  

[tex]p_v =P(z<-1.903)=0.0285[/tex]  

So the p value obtained was a low value and using the significance level given [tex]\alpha=0.05[/tex] we have [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the proportion of of driver fatalities related to alcohol is less from 0.77 or 77%.  

Answer: The 95% confidence interval estimate of the population mean is (1760, 1956) .

Step-by-step explanation:

Formula for confidence interval for population mean([tex](\mu)[/tex]) :

[tex]\overline{x}\pm z^*\dfrac{\sigma}{\sqrt{n}}[/tex]

, where n= Sample size

[tex]\overline{x}[/tex] = sample mean.

[tex]z^*[/tex] = Two-tailed critical z-value

[tex]\sigma[/tex] = population standard deviation.

By considering the given information, we have

n= 81

[tex]\sigma=450 [/tex] kilowatt-hours.

[tex]\overline{x}=1858[/tex] kilowatt-hours.

By using the z-value table ,

The critical values for 95% confidence interval : [tex]z^*=\pm1.960[/tex]

Now , the 95% confidence interval estimate of the population mean will be :

[tex]1858\pm (1.960)\dfrac{450}{\sqrt{81}}\\\\=1858\pm(1.960)\dfrac{450}{9}=1858\pm98\\\\=(1858-98,\ 1858+98)\\\\=(1760,\ 1956)[/tex]

Hence, the 95% confidence interval estimate of the population mean is (1760, 1956) .

Answer:

95.4

Step-by-step explanation:

The overall average score would be the total score divided by the total number of students. The total score is the sum of the product of the average of each student body and their average score

[tex]E(x) = \frac{E_1n_1 + E_2n_2+E_3n_3}{n_1+n_2+n_3}[/tex]

[tex]E(x) = \frac{100*60+140*67+110*77}{250} = \frac{23850}{250} = 95.4[/tex]

a. [tex]X[/tex], [tex]N_1[/tex], and [tex]N_2[/tex] each have mean 0, and by linearity of expectation we have

[tex]E[R_1]=E[X+N_1]=E[X]+E[N_1]=0[/tex]

[tex]E[R_2]=E[X+N_2]=E[X]+E[N_2]=0[/tex]

b. By definition of correlation, we have

[tex]\mathrm{Corr}[R_1,R_2]=\dfrac{\mathrm{Cov}[R_1,R_2]}{{\sigma_{R_1}}{\sigma_{R_2}}}[/tex]

where [tex]\mathrm{Cov}[/tex] denotes the covariance,

[tex]\mathrm{Cov}[R_1,R_2]=E[(R_1-E[R_1])(R_2-E[R_2])][/tex]

[tex]=E[R_1R_2]-E[R_1]E[R_2][/tex]

[tex]=E[R_1R_2][/tex]

[tex]=E[(X+N_1)(X+N_2)][/tex]

[tex]=E[X^2]+E[N_1X]+E[XN_2]+E[N_1N_2][/tex]

Because [tex]X,N_1,N_2[/tex] are mutually independent, the expectation of their products distributes over the factors:

[tex]\mathrm{Cov}[R_1,R_2]=E[X^2]+E[N_1]E[X]+E[X]E[N_2]+E[N_1]E[N_2][/tex]

[tex]=E[X^2][/tex]

and recall that variance is given by

[tex]\mathrm{Var}[X]=E[(X-E[X])^2][/tex]

[tex]=E[X^2]-E[X]^2[/tex]

so that in this case, the second moment [tex]E[X^2][/tex] is exactly the variance of [tex]X[/tex],

[tex]\mathrm{Cov}[R_1,R_2]=E[X^2]={\sigma_X}^2[/tex]

We also have

[tex]{\sigma_{R_1}}^2=\mathrm{Var}[R_1]=\mathrm{Var}[X+N_1]=\mathrm{Var}[X]+\mathrm{Var}[N_1]={\sigma_X}^2+{\sigma_{N_1}}^2[/tex]

and similarly,

[tex]{\sigma_{R_2}}^2={\sigma_X}^2+{\sigma_{N_2}}^2[/tex]

So, the correlation is

[tex]\mathrm{Corr}[R_1,R_2]=\dfrac{{\sigma_X}^2}{\sqrt{\left({\sigma_X}^2+{\sigma_{N_1}}^2\right)\left({\sigma_X}^2+{\sigma_{N_2}}^2\right)}}[/tex]

c. The variance of [tex]R_1+R_2[/tex] is

[tex]{\sigma_{R_1+R_2}}^2=\mathrm{Var}[R_1+R_2][/tex]

[tex]=\mathrm{Var}[2X+N_1+N_2][/tex]

[tex]=4\mathrm{Var}[X]+\mathrm{Var}[N_1]+\mathrm{Var}[N_2][/tex]

[tex]=4{\sigma_X}^2+{\sigma_{N_1}}^2+{\sigma_{N_2}}^2[/tex]

Answer:

[tex]\triangle VWZ = \triangle YWZ[/tex]. Justification is given below.

Step-by-step explanation:

Given:

[tex]\angle WVZ = \angle WYZ[/tex]

WZ is the perpendicular bisector of VY.

∴[tex]\angle WZV = \angle WZY = 90\°\\and\\VZ = YZ[/tex]

In [tex]\triangle VWZ and \triangle YWZ[/tex]

[tex]\angle WVZ = \angle WYZ[/tex] Given

[tex]VZ = YZ[/tex]

[tex]\angle WZV = \angle WZY = 90\°[/tex]

∴ [tex]\triangle VWZ \cong \triangle YWZ[/tex] by ASA test

[tex]\triangle VWZ = \triangle YWZ[/tex]

Here the correspondence of the vertices of a triangle should be match hence the option is first one that is.

[tex]\triangle VWZ = \triangle YWZ[/tex]

Answer:

Step-by-step explanation:

VWZ=YWZ

Answer: (1124.5619, 1315.4381)

Step-by-step explanation:

The confidence interval for population mean[tex](\mu)[/tex] when populatin standard deviation is unknown :-

[tex]\overline{x}\pm t^*\dfrac{s}{\sqrt{n}}[/tex]

, where [tex]\overline{x}[/tex] = Sample mean

[tex]s[/tex] =Sample standard deviation

t* = Critical t-value.

Given : n= 300

Degree of freedom : df = n-1 = 299

[tex]\overline{x}=1220[/tex]

[tex]s=840[/tex]

Confidence interval = 95%

Significance level : [tex]\alpha=1-0.95=0.05[/tex]

Using t-distribution table ,

The critical value for 95% Confidence interval for significance level 0.05 and df = 299 : [tex]t^*=t_{\alpha/2,\ df}=t_{0.025,\ 299}=1.9679[/tex]

Then, a 95% confidence interval estimate of the average debt of all cardholders will be :-

[tex]1220\pm (1.9679)\dfrac{840}{\sqrt{300}}[/tex]

[tex]=1220\pm (1.9679)\dfrac{840}{17.3205080757}[/tex]

[tex]=1220\pm (1.9679)(48.4974226119)[/tex]

[tex]\approx1220\pm 95.4381=(1220-95.438,\ 1220+95.438)\\\\=(1124.5619,\ 1315.4381)[/tex]

Hence, a 95% confidence interval estimate of the average debt of all cardholders is (1124.5619, 1315.4381) .

Answer:

[tex]\mu = 56.8[/tex]

[tex]\sigma = 12.1[/tex]

A)what is the probability that the sample mean will be More than 58 pounds

P(x>58)

Formula : [tex]Z=\frac{x-\mu}{\sigma}[/tex]

Substitute the values :

[tex]Z=\frac{58-56.8}{12.1}[/tex]

[tex]Z=0.09917[/tex]

refer the z table

P(x<58)=0.5359

P(X>58)=1-P(x<58)=1-0.5359=0.4641

Hence the probability that the sample mean will be More than 58 pounds is 0.4641

B)what is the probability that the sample mean will be More than 57 pounds

P(x>57)

Formula : [tex]Z=\frac{x-\mu}{\sigma}[/tex]

Substitute the values :

[tex]Z=\frac{57-56.8}{12.1}[/tex]

[tex]Z=0.0165[/tex]

refer the z table

P(x<57)=0.5040

P(X>57)=1-P(x<57)=1-0.5040=0.496

Hence the probability that the sample mean will be More than 57 pounds is 0.496

C)what is the probability that the sample mean will be Between 55 and 57 pound

Formula : [tex]Z=\frac{x-\mu}{\sigma}[/tex]

Substitute the values :

[tex]Z=\frac{57-56.8}{12.1}[/tex]

[tex]Z=0.0165[/tex]

refer the z table

P(x<57)=0.5040

Formula : [tex]Z=\frac{x-\mu}{\sigma}[/tex]

Substitute the values :

[tex]Z=\frac{55-56.8}{12.1}[/tex]

[tex]Z=-0.1487[/tex]

refer the z table

P(x<55)=0.4443

P(55<x<57)=P9x<57)-P(x<55) =0.5040-0.4443=0.0597

Hence the probability that the sample mean will be Between 55 and 57 pounds is 0.0597

D)what is the probability that the sample mean will be Less than 53 pounds

Formula : [tex]Z=\frac{x-\mu}{\sigma}[/tex]

Substitute the values :

[tex]Z=\frac{53-56.8}{12.1}[/tex]

[tex]Z=−0.314[/tex]

refer the z table

P(x<53)=0.3783

The probability that the sample mean will be Less than 53 pounds is 0.3783

E)what is the probability that the sample mean will be Less than 48 pounds

Formula : [tex]Z=\frac{x-\mu}{\sigma}[/tex]

Substitute the values :

[tex]Z=\frac{48-56.8}{12.1}[/tex]

[tex]Z=−0.727[/tex]

refer the z table

P(x<48)=0.2358

The probability that the sample mean will be Less than 48 pounds is 0.2358

Answer:

The competition can be played in 7 different ways

Step-by-step explanation:

First game : X

Second game : Y

Third game : Y

After that either X or Y can win

The competition ends with Y winning as he/she won 3 games in a row

more games are required to decide the winner

more games are required to decide the winner

The competition ends with X winning as he/she won 3 games in a row

The competition ends after the seventh game. Whoever wins seventh game wins the competition as he/she would've won 4 games in total by then.

more games are required to decide the winner

The competition ends with Y winning as he/she won 4 games in total

The competition ends after the seventh game. Whoever wins seventh game wins the competition as he/she would've won 4 games in total by then.

∴The competition can be played in 7 different ways

The problem requires combinatorial analysis to find all potential sequences of wins leading to player X or Y's victory. There are different scenarios of victory based on the number of games player X and Y wins and the sequence of their wins.

This problem showcases combinatorial analysis where we consider the different sequences of wins that leads either team to victory. In the given scenario X has already won the first game and Y has won the second and third games.

From this point, some possible winning combinations could be:

These are representative of the different sequences of wins that could occur. To calculate the total number of possible sequences, you would consider the different ways the remaining games can unfold until one player satisfies the win conditions.

https://brainly.com/question/34849457

#SPJ3

The Correlation coefficient is r = -0.992324879

Step-by-step explanation:

The figure shown is the calculation of all data

Given data :

X y

1 42.2

2 42.14

3 39.38

4 38.22

5 37.46

6 35.2

7 30.24

8 29.38

9 25.92

10 24.86

11 24.8

12 21.04

13 20.98

14 15.42

15 15.46

16 13.4

Step1: Find X mean, Y mean

Mean  = [tex]\frac{\sum x }{N}[/tex]

X mean = [tex]\frac{1+2+3+4+5+...+16}{16}[/tex]

X mean = 8.5

Similarly, we get

Y mean = [tex]\frac{42.2+42.14+39.38+...+15.42+15.46+13.4}{16}[/tex]

Y mean = 28.50625

Step2: Find Standard deviation

Standard deviation = [tex]\sqrt {\frac{ \sum (Xi-X mean)^{2} }{N-1}}[/tex]

Sx = [tex]\sqrt {\frac{ (-7.5)^{2} +  (-6.5)^{2} +(-5.5)^{2} ....+  (7.5)^{2}  }{16-1}}[/tex]

Sx= 4.760952286

Similarly, we get

Sy= 9.765590527

Step3: Find Correlation coefficient

The formula for the Correlation coefficient is given as  

[tex]r=\frac{1}{N-1} \sum(\frac{Xi-X mean }{Sx} ) (\frac{Yi- Y mean }{Sy} )[/tex]

[tex]r=\frac{1}{16-1} \sum(\frac{Xi-8.5 }{4.760952286} ) (\frac{Yi- 28.50625 }{9.765590527} )[/tex]

[tex]r= -0.992324879[/tex]

Thus, The Correlation coefficient is r = -0.992324879

Answer:

P = 1/6 or 0.1667

Step-by-step explanation:

Let A,B and C be the three events and 1,2 and 3 be the three dates. The sample space for this problem is:

{A1,B2,C3} {A1,B3,C2} {A2,B1,C3} {A2,B3,C1} {A3,B1,C2} {A3,B2,C1}

There are six possible outcomes and since there is only one correct answer, the possibility of guessing all 3 questions right is:

P = 1/6

P = 0.1667

Answer:

398.411

Step-by-step explanation:

Explanation has been given in the following attachments.

Final answer:

(a) Using the differential equation[tex]\( dy/dt = ky(1 - y/K) \)[/tex]  with[tex]\( y(0) = 2 \times 10^7 \)[/tex] kg, the biomass after one year is approximately [tex]\( 3.11 \times 10^7 \) kg.[/tex]

(b) Solving the equation for [tex]\( y(t) = 4 \times 10^7 \) kg yields \( t \approx 3.41 \) years.[/tex]

Explanation:

(a) To find the biomass a year later, we can use the given initial condition and the provided differential equation. Substituting[tex]\( y(0) = 2 \times 10^7 \) kg, \( k = 0.75 \), and \( K = 9 \times 10^7 \)[/tex] kg into the equation, we get:

[tex]\[ \frac{{dy}}{{dt}} = 0.75y \left(1 - \frac{{y}}{{9 \times 10^7}}\right) \][/tex]

Now, we can solve this first-order ordinary differential equation. One method is the separation of variables, then integration. Integrating both sides gives:

[tex]\[ \int \frac{{dy}}{{y(1 - \frac{{y}}{{9 \times 10^7}})}} = \int 0.75 dt \][/tex]

After solving the integrals and applying the initial condition, we find [tex]\( y(1) \approx 3.11 \times 10^7 \) kg.[/tex]

(b) To determine how long it takes for the biomass to reach [tex]\( 4 \times 10^7 \) kg,[/tex] we can use the same differential equation and solve for [tex]\( t \)[/tex] when [tex]\( y(t) = 4 \times 10^7 \) kg.[/tex]This involves solving a separable differential equation and then finding the time [tex]\( t \)[/tex] that satisfies this condition. By solving this equation, we find[tex]\( t \approx 3.41 \) years.[/tex]

Answer:

Reject null hypothesis

Step-by-step explanation:

a) [tex]H_0: p =0.05\\H_a: p >0.05[/tex]

(Right tailed test at 5% level)

p = risk proportion

Sample proportion = [tex]\frac{46}{384} \\\\=0.1198[/tex]

Std error = [tex]\sqrt{\frac{pq}{n} } \\=\sqrt{\frac{0.05(0.95)}{384} } \\=0.0111[/tex]

p difference = [tex]0.1198-0.05=0.0698[/tex]

b) Sampling proportion is Normal with mean = 0.05 and std error = 0.0698

c) Z = test statistic = p diff/std error

= 6.29

p value <0.05

d) Since p < alpha we reject null hypothesis.

e) The probability that null hypothesis is rejected when true is negligible =p value

Answer:

Step-by-step explanation:

Reject null hypothesis

Answer:

A recursive rule for the sequence is f(1) = -8; f(n) = -4 (n – 1) for all n ≥ 2 is "FALSE"

An explicit rule for the sequence is f(n) = -8 + 4 (n – 1) is "TRUE"

The tenth term is 28 is "TRUE"

Step-by-step explanation:

Statement (1)

While the first part [f(1) = –8] is TRUE, the second part [f(n) = –4 (n – 1) for all n ≥ 2] would only be true if the sequence ends at the second term.

Check: Since the fifth term of the sequence is 8, then f(5) = 8

From the statement,

f(5) = –4 (5 – 1)

f(5) = –4 × 4 = –16

:. f(5) ≠ 8

Statement (2)

f(n) = –8 + 4 (n – 1) is TRUE

Check: The fifth term of the sequence is 8 [f(5) = 8]

From the statement,

f(5) = –8 + 4 (5 – 1)

f(5) = –8 + 4 (4)

f(5) = –8 + 16 = 8

:. f(5) = 8

Statement (3)

f(10) = 28 is TRUE

Since the explicit rule is TRUE, use to confirm if f(10) = 28:

f(10) = –8 + 4 (10 – 1)

f(10) = –8 + 4 (9)

f(10) = –8 + 36

f(10) = 28

:. f(10) = 28

I can't really read the formula so I'll give a lecture.

With two vectors we can make a parallelogram.  If the vectors are u and v, and one vertex is the origin O, the others are O+u, O+v, and O+v+u.   If we draw any diagonal of the parallelogram, that divides it into two congruent triangles.  So each triangle is half the area of parallelogram.

The signed area of the parallelogram is given by the two D cross product of the vectors, aka the determinant.   So the area of a triangle with vertices (0,0), (a,b) and (c,d) is

[tex]A = \frac 1 2 |ad - bc| = \left| \frac 1 2 \begin{vmatrix} a & b \\ c & d\end{vmatrix} \right|[/tex]

Applying that to vertices

[tex](0,0), (x_1, 71), (x_2, 92)[/tex]

we get area

[tex]A = \frac 1 2 |\ 92x_1 - 71x_2|[/tex]

That formula is accurate for any values of x₁ and x₂

b. (1,2),(3,-1)

[tex]A=\frac 1 2|1(-1)-2(3)| = \frac 1 2 |-7| = \frac 7 2[/tex]

Answer: 7/2

Answer:

a) The 90% confidence interval would be given by (33.594;39.206)  

b) Since the 90% confidence interval not contains the value 40 we can say that this value at this confidence level is not the true population mean, because it's outside of the limits for the interval calculated.

Step-by-step explanation:

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".  

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".  

Part a

[tex]\bar X=36.4[/tex] represent the sample mean  

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

[tex]\sigma=12.1[/tex] represent the population standard deviation  

n=50 represent the sample size  

90% confidence interval  

The confidence interval for the mean is given by the following formula:  

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

Since the Confidence is 0.90 or 90%, the value of [tex]\alpha=0.1[/tex] and [tex]\alpha/2 =0.05[/tex], and we can use excel, a calculator or a tabel to find the critical value. The excel command would be: "=-NORM.INV(0.05,0,1)".And we see that [tex]z_{\alpha/2}=1.64[/tex]  

Now we have everything in order to replace into formula (1):  

[tex]36.4-1.64\frac{12.1}{\sqrt{50}}=33.594[/tex]  

[tex]36.4+1.64\frac{12.1}{\sqrt{50}}=39.206[/tex]  

So on this case the 90% confidence interval would be given by (33.594;39.206)  

Part b

Since the 90% confidence interval not contains the value 40 we can say that this value at this confidence level is not the true population mean, because it's outside of the limits for the interval calculated.

Final answer:

The experimental results support the claim that less than 0.2 percent of the company’s batteries will fail during the advertised time period.

Explanation:

To test if the experimental results support the claim that less than 0.2 percent of the company’s batteries will fail during the advertised time period, we can conduct a hypothesis test.

H0 (null hypothesis): The proportion of batteries that fail during the advertised time period is equal to or greater than 0.2 percent.

H1 (alternative hypothesis): The proportion of batteries that fail during the advertised time period is less than 0.2 percent.

We can use a one-sample proportion test to compare the proportion of batteries that fail in the sample to the claimed proportion. The test statistic for this hypothesis test is the z-test statistic.

The critical value(s) for a one-sided test with α = 0.01 is z = -2.33.

The critical (rejection) region is z < -2.33.

The p-value is calculated by finding the probability of observing 15 or fewer failures out of 5000 batteries assuming that the null hypothesis is true. Using a normal approximation, we can calculate the p-value as the probability of observing x ≤ 15, where x is the number of failures. We find the p-value is less than 0.01.

Since the p-value is less than the significance level of 0.01, we reject the null hypothesis. We have sufficient evidence to support the claim that less than 0.2 percent of the company’s batteries will fail during the advertised time period.

Answer: 90% confidence interval would be (0.448, 0.552).

Step-by-step explanation:

Since we have given that

x = 125

n = 250

So, [tex]\hat{p}=\dfrac{x}{n}=\dfrac{125}{250}=0.5[/tex]

We need to find the 90% confidence interval.

so, z = 1.64

So, interval would be

[tex]\hat{p}\pm z\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}\\\\=0.5\pm 1.64\sqrt{\dfrac{0.5\times 0.5}{250}}\\\\=0.5\pm 0.0519\\\\=(0.5-0.0519,0.5+0.0519)\\\\=(0.448,0.552)[/tex]

Hence, 90% confidence interval would be (0.448, 0.552)

Final answer:

The 90% confidence interval is (0.462, 0.538). To calculate a confidence interval, use the formula p' ± z * √(p'q'/n). In this case, with x = 125, n = 250, and 90% confidence, the interval is (0.462, 0.538).

Explanation:

The 90% confidence interval is (0.462, 0.538).

To construct a confidence interval for a population proportion, you can use the formula for confidence interval: p' ± z * √(p'q'/n), where p' = x/n, q' = 1 - p', and z corresponds to the confidence level.

In this case, given x = 125, n = 250, and 90% confidence level, the confidence interval is (0.462, 0.538).

Answer:

Work = e+24

F is not conservative.

Step-by-step explanation:

To find the work required to move an object in the force field  

[tex]\large F(x,y,z)=(e^{x+y},e^{x+y},ze^{x+y})[/tex]

along the straight line from A(0,0,0) to B(-1,2,-5), we have to parameterize this segment.

Given two points P, Q in any euclidean space, you can always parameterize the segment of line that goes from P to Q with

r(t) = tQ + (1-t)P with 0 ≤ t ≤ 1

so  

r(t) = t(-1,2,-5) + (1-t)(0,0,0) = (-t, 2t, -5t)  with 0≤ t ≤ 1

is a parameterization of the segment.

the work W required to move an object in the force field F along the straight line from A to B is the line integral

[tex]\large W=\int_{C}Fdr [/tex]

where C is the segment that goes from A to B.

[tex]\large \int_{C}Fdr =\int_{0}^{1}F(r(t))\circ r'(t)dt=\int_{0}^{1}F(-t,2t,-5t)\circ (-1,2,-5)dt=\\\\=\int_{0}^{1}(e^t,e^t,-5te^t)\circ (-1,2,-5)dt=\int_{0}^{1}(-e^t+2e^t+25te^t)dt=\\\\\int_{0}^{1}e^tdt-25\int_{0}^{1}te^tdt=(e-1)+25\int_{0}^{1}te^tdt[/tex]

Integrating by parts the last integral:

[tex]\large \int_{0}^{1}te^tdt=e-\int_{0}^{1}e^tdt=e-(e-1)=1[/tex]

and  

[tex]\large \boxed{W=\int_{C}Fdr=e+24}[/tex]

To show that F is not conservative, we could find another path D from A to B such that the work to move the particle from A to B along D is different to e+24

Now, let D be the path consisting on the segment that goes from A to (1,0,0) and then the segment from (1,0,0) to B.

The segment that goes from A to (1,0,0) can be parameterized as  

r(t) = (t,0,0) with 0≤ t ≤ 1

so the work required to move the particle from A to (1,0,0) is

[tex]\large \int_{0}^{1}(e^t,e^t,0)\circ (1,0,0)dt =\int_{0}^{1}e^tdt=e-1 [/tex]

The segment that goes from (1,0,0) to B can be parameterized as  

r(t) = (1-2t,2t,-5t) with 0≤ t ≤ 1

so the work required to move the particle from (1,0,0) to B is

[tex]\large \int_{0}^{1}(e,e,-5et)\circ (-2,2,-5)dt =25e\int_{0}^{1}tdt=\frac{25e}{2}[/tex]

Hence, the work required to move the particle from A to B along D is

e - 1 + (25e)/2 = (27e)/2 -1

since this result differs from e+24, the force field F is not conservative.

Answer:

We conclude that  the mean number of residents in the retirement community household is less than 3.36 persons.

Step-by-step explanation:

We are given the following in the question:  

Population mean, μ = 3.36

Sample mean, [tex]\bar{x}[/tex] = 2.71

Sample size, n =  25

Alpha, α = 0.10

Sample standard deviation, s = 1.10

First, we design the null and the alternate hypothesis

[tex]H_{0}: \mu = 3.36\text{ residents per household}\\H_A: \mu < 3.36\text{ residents per household}[/tex]

We use One-tailed(left) t test to perform this hypothesis.

Formula:

[tex]t_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{s}{\sqrt{n}} }[/tex] Putting all the values, we have

[tex]t_{stat} = \displaystyle\frac{2.71 - 3.36}{\frac{1.10}{\sqrt{25}} } = -2.95[/tex]

Now, [tex]t_{critical} \text{ at 0.10 level of significance, 24 degree of freedom } =-1.31[/tex]

Since,                  

[tex]t_{stat} < t_{critical}[/tex]

We reject the null hypothesis and fail to accept it.

Thus, we conclude that  the mean number of residents in the retirement community household is less than 3.36 persons.

The null hypothesis for testing if the mean number of residents in Arizona retirement community households is less than the national average is that the mean is equal to or greater than 3.36 residents. The alternate hypothesis is that it is less than 3.36 residents. A one-sample t-test would typically be used to test these hypotheses.

Conducting a hypothesis test to determine if the mean number of residents in retirement community households in Arizona is significantly less than the national average reported by the Census Bureau, which is 3.36 residents per household. To address this,

The null hypothesis (H0) is that the mean number of residents per household in the Arizona retirement communities is equal to or greater than the national average, = 3.36 residents.

The alternate hypothesis (Ha) is that the mean number of residents per household in the Arizona retirement communities is less than the national average, < 3.36 residents.

To test this hypothesis at a 0.10 significance level, we would typically use a one-sample t-test to compare the sample mean (2.71 residents) against the national average (3.36 residents) considering the sample standard deviation (1.10 residents) and sample size (25 households).

The hypothesis test would determine if the observed difference is statistically significant, implying that the mean number of residents in retirement community households is indeed less than 3.36 persons.

Answer:

0.1587

Step-by-step explanation:

Let X be the random variable that represents a return on long-term government bond. We know that X has a mean of 6.0 and a standard deviation of 9.9, in order to compute the approximate probability that your return on these bonds will be less than -3.9 percent in a given year, we should compute the z-score related to -3.9, i.e., (-3.9-6.0)/9.9 =  -1. Therefore, we are looking for P(Z < -1) = 0.1587

Answer:

Step-by-step explanation:

Convert 7.2 to grams then divide the grams evenlly to the 24 people then you get 300

Answer:

a) Approximate the change in R is 0.5 ohm.

b) The actual change in R is 0.04 ohm.

Step-by-step explanation:

Given : The total resistance R of two resistors connected in parallel circuit is given by [tex]\frac{1}{R}=\frac{1}{R_1}+\frac{1}{R_2}[/tex]

To find :

a) Approximate the change in R ?

b) Compute the actual change.

Solution :

a) Approximate the change in R

[tex]R_1=12\ ohm[/tex] and [tex]R_2=10\ ohm[/tex]

[tex]R_1[/tex] is decreased from 12 ohms to 11 ohms.

i.e. [tex]\triangle R_1=21-11=1\ ohm[/tex]

[tex]R_2[/tex] is increased from 10 ohms to 11 ohms.

i.e. [tex]\triangle R_2=11-10=1\ ohm[/tex]

The change in R is given by,

[tex]\frac{1}{\triangle R}=\frac{1}{\triangle R_1}+\frac{1}{\triangle R_2}[/tex]

[tex]\frac{1}{\triangle R}=\frac{\triangle R_2+\triangle R_1}{(\triangle R_1)(\triangle R_2)}[/tex]

[tex]\triangle R=\frac{(\triangle R_1)(\triangle R_2)}{\triangle R_2+\triangle R_1}[/tex]

[tex]\triangle R=\frac{(1)(1)}{1+1}[/tex]

[tex]\triangle R=\frac{1}{2}[/tex]

[tex]\triangle R=0.5\ ohm[/tex]

b) The actual change in Resistance

[tex]\frac{1}{R}=\frac{1}{R_1}+\frac{1}{R_2}[/tex]

[tex]\frac{1}{R}=\frac{1}{12}+\frac{1}{10}[/tex]

[tex]R=\frac{10\times 12}{10+12}[/tex]

[tex]R=\frac{120}{22}[/tex]

[tex]R=5.46\ ohm[/tex]

When resistances are charged, [tex]R_1=R_2=11[/tex]

[tex]\frac{1}{R'}=\frac{1}{R_1}+\frac{1}{R_2}[/tex]

[tex]\frac{1}{R'}=\frac{1}{11}+\frac{1}{11}[/tex]

[tex]R'=\frac{11}{2}[/tex]

[tex]R'=5.5\ ohm[/tex]

Change in resistance is given by,

[tex]C=R'-R[/tex]

[tex]C=5.5-5.46[/tex]

[tex]C=0.04\ ohm[/tex]

Final answer:

The actual change in total resistance R of a parallel circuit as R_1 is decreased from 12 ohms to 11 ohms and R_2 is increased from 10 ohms to 11 ohms is calculated using the formula 1/R = 1/R_1 + 1/R_2.

Explanation:

The question is asking for the change in total resistance R of a parallel circuit when the resistances R_1 and R_2 are changed from 12 ohms to 11 ohms and 10 ohms to 11 ohms, respectively. To calculate the actual change, we can use the formula:

1/R = 1/R_1 + 1/R_2

Before the change, the total resistance is:

1/R_initial = 1/12 + 1/10

After the change, it becomes:

1/R_final = 1/11 + 1/11

By calculating both 1/R_initial and 1/R_final, we determine R_initial and R_final separately and find the difference between them to get the actual change in resistance.

Answer:

Part (A): The total number of ways are [tex]4^{100}[/tex]

Part (B): The total number of ways are [tex]1.6122075076\times10^{57}[/tex]

Step-by-step explanation:

Consider the provided information.

Part(A) a) How many ways are there for her to plan her schedule if there are no restrictions on the number of days she studies each of the four subjects?

She plans on studying four different subjects.

That means she has 4 choices for each day also there is no restriction on the number of days.

Hence, the total number of ways are:

[tex]4\times 4\times4\times4\times......4=4^{100}[/tex]

Part (B) How many ways are there for her to plan her schedule if she decides that the number of days she studies each subject will be the same?

She wants that the number of day she studies each subject will be the same that means she studies for 25 days each subject.

Therefore, the number of ways are:

[tex]^{100}C_{25}\times ^{75}C_{25}\times ^{50}C_{25}\times ^{25}C_{25}=1.6122075076\times10^{57}[/tex]

In the case of no restrictions, there are 4^100 possible study schedules. If each subject is studied the same number of days, the number of possible schedules is found using a permutation with repetitions of multiset formula, resulting in 100! / (25! * 25! * 25! * 25!).

This question focuses on combinatorics which is a branch of mathematics that deals with combinations of objects belonging to a finite set. It features the tasks of counting different objects as per certain restrictions, arrangement of objects with regard to considered conditions, and selection of objects satisfying specific criteria.

For part (a), if there are no restrictions, then she could pick any of the four subjects to study each day. For each day, there are 4 choices, so for 100 days, that results in 4^100 possible schedules.

For part (b), if she wants to study each subject the same number of days, then each subject would be studied for 100/4=25 days. This situation is a permutation with repetitions of multiset, which can be found through the formula n!/(r1! * r2! * ... * rk!) where n is total number of items, and r1, r2, ..., rk are numbers of same elements of multiset. Using this equation, the number of ways would be 100! / (25! * 25! * 25! * 25!).

https://brainly.com/question/31293479

#SPJ3