the height of woman ages 20-29 is normally distributed , with a mean of 64.4 inches. assuming the standard diviation = 2.3 inches. are you more likely to randomly select 1 woman with a height less than 64.9 inches or are you more likely to select a sample of 22 woman with a mean height less than 64.9 inches.(USING STANDARD NORMAL TABLE)A. what is the probability of randomly selecting 1 woman with a height less than 64.9 inches?B. what is the probability of selecting a sample of 22 woman with a mean height less then 64.9 inches?C. are you more likely to randomly select 1 woman with a height less than 64.9 inches or are you more likely to select a sample of 22 woman with a mean height of 64.9 inches?

Answer:

The function y must be decreasing (or equal to 0) on any interval on which it is defined.

Step-by-step explanation:

The derivative of a function gives us the rate at which that function is changing. In this case, -y^2, yields a negative value for every possible value of y, thus, the rate of change is always negative and the function y is decreasing (or equal to 0) on any interval on which it is defined.

The differential equation y' = -[tex]y^2[/tex] implies that y is either decreasing or constant wherever it is defined, because the derivative y' is non-positive.

By examining the differential equation y' = -[tex]y^2[/tex], we can infer some characteristics about the solutions without solving it. If y is a solution to this equation, then y' represents the derivative of y with respect to x. This derivative tells us about the rate of change of the function y.

Since the right side of the equation is -[tex]y^2[/tex], and a square of a real number is always non-negative, multiplying by -1 makes it non-positive. This implies that the derivative y' is either less than or equal to zero. Therefore, wherever the function y is defined, it must be either decreasing or constant (equal to zero). If y is positive, y will decrease because of the negative sign in front of the square. If y is negative, squaring it results in a positive number, but the negative sign still ensures that the rate of change is non-positive.

Conclusion: the function y is decreasing or remains constant on any interval it is defined; it cannot be increasing.

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Answer:

1/2

Step-by-step explanation:

The interior of the square is the region D = { (x,y) : 0 ≤ x,y ≤1 }. We call L(x,y) = 7y²x, M(x,y) = 8x²y. Since C is positively oriented, Green Theorem states that

[tex]\int\limits_C {L(x,y)} \, dx + {M(x,y)} \, dy = \int\limits^1_0\int\limits^1_0 {(Mx - Ly)} \, dxdy[/tex]

Lets calculate the partial derivates of M and L, Mx and Ly. They can be computed by taking the derivate of the respective value, treating the other variable as a constant.

Thus, Mx(x,y) - Ly(x,y) = 2xy, and therefore, the line ntegral is equal to the double integral

[tex] \int\limits^1_0\int\limits^1_0 {2xy} \, dxdy[/tex]

We can compute the double integral by applying the Barrow's Rule, a primitive of 2xy under the variable x is x²y, thus the double integral can be computed as follows

[tex]\int\limits^1_0\int\limits^1_0 {2xy} \, dxdy = \int\limits^1_0 {x^2y} |^1_0 \,dy = \int\limits^1_0 {y} \, dy = \frac{y^2}{2} \, |^1_0 = 1/2[/tex]

We conclude that the line integral is 1/2

To evaluate the line integral using Green's Theorem, we need to find the curl of the vector field and the area enclosed by the square. The line integral of the vector field along the square is equal to the double integral of the curl over the region enclosed by the square. Using this method, we can find the value of the line integral to be 1/3.

To evaluate the line integral using Green's Theorem, we first need to find the curl of the vector field. In this case, the vector field is F(x, y) = 7y^2x i + 8x^2y j. Taking the partial derivatives of its components with respect to x and y, we get curl(F) = (8x^2 - 14xy^2) k.

Next, we need to find the area enclosed by the square C, which is 1 unit^2. Using Green's Theorem, the line integral of F along C is equal to the double integral of curl(F) over the region D enclosed by C. Integrating curl(F) with respect to y, we get -7xy^2 + 6x^2y. Integrating this with respect to x over the given limits, we find the value of the line integral to be 1/3.

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Answer:

y = 2x^2

Step-by-step explanation:

The the coordinates (2,8) are expressed in terms of (x,y). Substitute the x with 2 and y with 8 and plug them into the formulas. The one that has the two sides equal to each other is the function that has (2,8) on its line.

Answer:

a) [tex]P(X\leq 12)=0.8586[/tex]

b) [tex]P(X\geq 5)=0.9802[/tex]

c) [tex]P(5\leq X\leq 12)=0.8389[/tex]

Step-by-step explanation:

The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".

Let X the random variable of interest, on this case we now that:

[tex]X \sim Binom(n=20, p=0.48)[/tex]

The probability mass function for the Binomial distribution is given as:

[tex]P(X)=(nCx)(p)^x (1-p)^{n-x}[/tex]

Where (nCx) means combinatory and it's given by this formula:

[tex]nCx=\frac{n!}{(n-x)! x!}[/tex]

We need to check the conditions in order to use the normal approximation.

[tex]np=20*0.48=9.6 \approx 10 \geq 10[/tex]

[tex]n(1-p)=20*(1-0.48)=10.4 \geq 10[/tex]

So we see that we satisfy the conditions and then we can apply the approximation.

If we appply the approximation the new mean and standard deviation are:

[tex]E(X)=np=20*0.48=9.6[/tex]

[tex]\sigma=\sqrt{np(1-p)}=\sqrt{20*0.48(1-0.48)}=2.234[/tex]

Part a

We want this probability:

[tex]P(X\leq 12)[/tex]

We can use the z score given by this formula [tex]Z=\frac{x-\mu}{\sigma}[/tex].

[tex]P(X\leq 12)=P(\frac{X-\mu}{\sigma}\leq \frac{12-9.6}{2.234})=P(Z\leq 1.074)=0.8586[/tex]

Part b

We want this probability:

[tex]P(X\geq 5)[/tex]

We can use again the z score formula and we have:

[tex]P(X\geq 5)=1-P(X<5)=1-P(\frac{X-\mu}{\sigma}< \frac{5-9.6}{2.234})=1-P(Z<- 2.059)=0.9802[/tex]

Part c

We want this probability:

[tex]P(5\leq X\leq 12)=P(\frac{5-9.6}{2.234}\leq \frac{X-\mu}{\sigma}\leq \frac{12-9.6}{2.234})=P(-2.059\leq Z \leq 1.074)[/tex]

[tex]=P(Z<1.074)-P(Z<-2.059)=0.8586-0.0197=0.8389[/tex]

Answer:

Answer is 13.3 ft

Step-by-step explanation:

i explained in the image below

Answer:

absolute max= (4.243,18)

absolute min =(-1,-5.916)

absolute max=(pi/6, 2.598)

absolute min = (pi/2,0)

Step-by-step explanation:

a) [tex]f(t) = t\sqrt{36-t^2} \\[/tex]

To find max and minima in the given interval let us take log and differentiate

[tex]log f(t) = log t + 0.5 log (36-t^2)\\Y(t) = log t + 0.5 log (36-t^2)[/tex]

It is sufficient to find max or min of Y

[tex]y'(t) = \frac{1}{t} -\frac{t}{36-t^2} \\\\y'=0 gives\\36-t^2 -t^2 =0\\t^2 =18\\t = 4.243,-4.243[/tex]

In the given interval only 4.243 lies

And we find this is maximum hence maximum at  (4.243,18)

Minimum value is only when x = -1 i.e. -5.916

b) [tex]f(t) = 2cost +sin 2t\\f'(t) = -2sint +2cos2t\\f"(t) = -2cost-4sin2t\\[/tex]

Equate I derivative to 0

-2sint +1-2sin^2 t=0

sint = 1/2 only satisfies I quadrant.

So when t = pi/6 we have maximum

Minimum is absolute mini in the interval i.e. (pi/2,0)

Answer: [tex]60,540< \mu<70,060[/tex]

Step-by-step explanation:

The confidence interval for population mean is given by :-

[tex]\overline{x}-z^*\dfrac{\sigma}{\sqrt{n}}< \mu<\overline{x}+z^*\dfrac{\sigma}{\sqrt{n}}[/tex]

, where [tex]\sigma[/tex] = Population standard deviation.

n= sample size

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

z* = Critical z-value .

Given :  [tex]\sigma=\$17,000[/tex]

n= 49

[tex]\overline{x}= \$65,300[/tex]

Two-tailed critical value for 95% confidence interval = [tex]z^*=1.960[/tex]

Then, the 95% confidence interval would be :-

[tex]65,300-(1.96)\dfrac{17000}{\sqrt{49}}< \mu<65,300+(1.96)\dfrac{17000}{\sqrt{49}}[/tex]

[tex]=65,300-(1.96)\dfrac{17000}{7}< \mu<65,300+(1.96)\dfrac{17000}{7}[/tex]

[tex]=65,300-4760< \mu<65,300+4760[/tex]

[tex]=60,540< \mu<70,060[/tex]

Hence, the 95​% confidence interval for estimating the population mean [tex](\mu)[/tex] :

[tex]60,540< \mu<70,060[/tex]

Final answer:

To test if there has been a significant decrease in the average price of homes sold in the U.S., we will conduct a hypothesis test using the p-value approach. We will define the null and alternative hypotheses, calculate the Z-test statistic, compare the p-value to the significance level, and make a conclusion based on the results.

Explanation:

To test if there has been a significant decrease in the average price of homes sold in the U.S., we will conduct a hypothesis test using the p-value approach. Let's define our hypotheses:

Null hypothesis (H0): The average price of homes sold in the U.S. is not significantly different from $220,000.

Alternative hypothesis (Ha): The average price of homes sold in the U.S. is significantly less than $220,000.

Next, we need to calculate the test statistic. Since the population standard deviation is known, we can use a Z-test. The formula for the Z-test statistic is:

Z = (sample mean - population mean) / (population standard deviation / sqrt(sample size))

In this case, the sample mean is $210,000, the population mean is $220,000, the population standard deviation is $36,000, and the sample size is 81. Plugging these values into the formula:

Z = (210,000 - 220,000) / (36,000 / sqrt(81))

Calculating this gives us a Z-statistic of -1.6875.

The final step is to compare the p-value of the test statistic to the significance level. Since the significance level is 1%, the critical value is 0.01. We can use a Z-table or calculator to find the p-value corresponding to a Z-statistic of -1.6875. The p-value is approximately 0.0466.

Since the p-value is less than the significance level of 0.01, we reject the null hypothesis. This means that there is sufficient evidence to conclude that there has been a significant decrease in the average price of homes sold in the U.S.

Answer:

B

Step-by-step explanation:

Answer:

sin(t)j - cos(t)i

Step-by-step explanation:

Let's start with A:

Position vector = cos(t)i

Velocity vector = -sin(t)i (differentiating the position vector)

acceleration vector = -cos(t)i   (differentiating the velocity vector)

Then we go to B:

Position vector = sin(t)j

Velocity vector = cos(t)j

acceleration vector = -sin(t)j

Relative acceleration = -cos(t)i - (-sin(t)j) = sin(t)j - cos(t)i

Answer:

a)  13/52    and the second  12/51

b) Two solutions:

b.1 if we did not picked up an eight in the first two cards   4/50

b.2 there is an eight i the two previous    3/59

c ) (48/52)*(47/51)*(46/50)

Step-by-step explanation:

Condition: Cards are taken out without replacement

a) Probability of first card is heart

There are 52 cards and 4 suits with the same probability , so you can compute this probability in two ways

we have  13 heart cards and 52 cards  then probability of one heart card is  13/52   = 0.25

or you have 4 suits, to pick up one specific suit the probability is 1/4 = 0,25

Now we have a deck of 51 card with 12 hearts, the probability of take one heart is : 12/51

b) There are 4 eight (one for each suit )   P =  4/50 if neither the first nor the second card was an eight of heart, if in a) previous we had an eight, then this probability change to 3/50

c) The probability of the first card different from an ace is 48/52 , the probability of the second one different of an ace is 47/51 and for the thirsd card is 46/50. The probability of none of the three cards is an ace is

(48/52)*(47/51)*(46/50)

Answer:

a) t =  2.7638

b) t = - 2.6245

c) t = 3.1058      on the right side    and

   t  = -3.1058    on the left

Step-by-step explanation:

a)Determine critical value for a right-tail test  for α = 0.01 level of significance and 10 degrees of fredom

From t-student table we find:

gl  =  10   and α = 0.01       ⇒  t =  2.7638

b)Determine   critical value for a left-tail test  for α = 0.01 level of significance  and sample size n = 15

From t-student table we find:

gl  =  14   and α = 0.01           gl  =  n - 1      gl  = 15 - 1    gl = 14

t = - 2.6245

c) Determine critical value for a two tails-test  for α = 0.01 level of  significance the  α/2   =   0.005  and sample size  n = 12

Then

gl  =  11   and α = 0.005

t = 3.1058      on the right side of the curve and by symmetry

t = - 3.1058  

From t-student table we find:

Answer:

The answer is a 100,000-choose-99 (a combination.)

[tex]P(100000, 99) = \displaystyle \left( \begin{array}{c}100000\cr 99\end{array}\right)[/tex].

Step-by-step explanation:

A combination [tex]C(n, r)[/tex] or equivalently [tex]\displaystyle \left(\begin{array}{c}n \cr r\end{array}\right)[/tex] gives the number of ways to choose [tex]r[/tex] out of [tex]n[/tex] elements.

A permutation [tex]P(n, r)[/tex] also gives the number of ways to choose [tex]r[/tex] out of [tex]n[/tex] elements. On top of that, it accounts for the order of the elements. Two elements in different order counts twice in a permutation, but only once in a combination.

The question emphasize that the council members are "co-equal." That implies that the order of the members don't really matter. Hence a combination with [tex]n = 100000[/tex] and [tex]r = 99[/tex] would be a more suitable choice.

Answer: 95% confidence interval would be (0.344,0.456).

Step-by-step explanation:

Since we have given that

n = 295

x = 118

so, [tex]\hat{p}=\dfrac{x}{n}=\dfrac{118}{295}=0.4[/tex]

At 95% confidence, z = 1.96

So, margin of error would be

[tex]z\times \sqrt{\dfrac{p(1-p)}{n}}\\\\=1.96\times \sqrt{\dfrac{0.4\times 0.6}{295}}\\\\=0.056[/tex]

so, 95% confidence interval would be

[tex]\hat{p}\pm \text{margin of error}\\\\=0.4\pm 0.056\\\\=(0.4-0.056,0.4+0.056)\\\\ =(0.344,0.456)[/tex]

Hence, 95% confidence interval would be (0.344,0.456).

Answer:

Step-by-step explanation:

Answer:

± 0.0736

Step-by-step explanation:

Data provided in the question:

randomly chosen graduates of California medical schools last year intended to specialize in family practice, p = [tex]\frac{48}{120}[/tex] = 0.4

Confidence level = 90%

sample size, n = 120

Now,

For 90% confidence level , z-value = 1.645

Width of the confidence interval = ± Margin of error

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

= ± [tex]1.645\times\sqrt\frac{0.4\times(1-0.4)}{120}[/tex]

= ± 0.07356 ≈ ± 0.0736

Hence,

The correct answer is option  ± 0.0736

The correct option is a. [tex]\±0.0736.[/tex] The width of a [tex]90\%[/tex] confidence interval for the proportion that plan to specialize in family practice

To find the width of a confidence interval for a proportion, we can use the formula:

[tex]\[ \text{Width} = Z \times \sqrt{\frac{p(1-p)}{n}} \][/tex]

where:

[tex]\( Z \)[/tex] is the Z-score corresponding to the desired confidence level,

[tex]\( p \)[/tex] is the sample proportion,

[tex]\( n \)[/tex] is the sample size.

Given:

[tex]Sample\ proportion \( p = \frac{48}{120} = \frac{2}{5} = 0.4 \)[/tex],

[tex]Sample\ size \( n = 120 \)[/tex]

[tex]Confidence \ level = 90\%[/tex], which corresponds to a Z-score of approximately [tex]1.645.[/tex]

Substitute the values into the formula:

[tex]\[ \text{Width} = 1.645 \times \sqrt{\frac{0.4 \times 0.6}{120}} \][/tex]

[tex]\[ \text{Width} = 1.645 \times \sqrt{\frac{0.24}{120}} \][/tex]

[tex]\[ \text{Width} = 1.645 \times \sqrt{0.002} \][/tex]

[tex]\[ \text{Width} = 1.645 \times 0.0447 \][/tex]

[tex]\[ \text{Width} = 0.0736 \][/tex]

Answer:

53¢

Step-by-step explanation:

First, I'll put these in order.

20¢;30¢; 30¢;75¢;40¢;40¢;40¢;40¢;50¢;55¢55¢65¢;65¢; $1.50;  

Then, I'll combine like terms.

30+30=60

40+40+40+40(or 40 x 4)=160

55+55=110

65+65=130

60+160+110+130+20+75+50+$1.50=$7.55/14=53¢

PLZ correct me if i'm wrong :-D

Answer:

5/8,-5√2,{(2x+3)(x-5)}

Step-by-step explanation:

1) 5x-10/8x-16

=5(x-2)/8(x-2)

=5/8

2) √32-3√18

=4√2-3√18

=4√2-9√2

=(4-9)√2

= -5√2

4) 2x^2-7x-15

=2x^2+3x-10x-15

=(2x+3),(x-5)

Answer:

Option D is right

Step-by-step explanation:

given that a two sample test for mean of independent samples to be done.

We create hypotheses as:

[tex]H_0: \ bar x = \bar y[/tex] vs alternate suitably right or left or two tailed according to the needs.

The conditions needed for conducting this test would be

I. Both populations are approximately normally distributed

II. Both sample sizes greater than 30

III. Population of differences is approximately normally distributed

i.e. either i, ii or III

Option D is right.

Answer:

a) Nina decreases the confidence level to 90%?  (Decrease)

b) Nina decreases the sample size to 34 locations? (Increase)

c) Nina increases the sample size to 70 locations?   (Decrease)

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".

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

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

s represent the sample standard deviation

n=48 represent the original sample size  

Confidence =95% or 0.95

ME=4.28 represent the margin of error.

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

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

In order to calculate the mean and the sample deviation we can use the following formulas:  

[tex]\bar X= \sum_{i=1}^n \frac{x_i}{n}[/tex] (2)  

[tex]s=\sqrt{\frac{\sum_{i=1}^n (x_i-\bar X)}{n-1}}[/tex] (3)  

And the margin of error is given by the following expression:

[tex]ME= t_{\alpha/2}\frac{s}{\sqrt{n}}[/tex]   (4)

Based on the formula (4) we can answer all the questions involved:

a) Nina decreases the confidence level to 90%?

On this case the value for [tex]t_{\alpha/2}[/tex] will also decrease so the margin of error would decrease.

b) Nina decreases the sample size to 34 locations?

If we analyze the original sample size of 48 we see that if we reduce the value of n to 34, the margin of error would increase, because n is on the denominator of the margin of error.

c) Nina increases the sample size to 70 locations?

If we analyze the original sample size of 48 we see that if we increase the value of n to 70, the margin of error would decrease, because n is on the denominator of the margin of error.

Answer:

209 points

Step-by-step explanation:

Mean points scored (μ) = 182 points  

Standard deviation (σ) = 14 points

The z-score for any given game score 'X' is defined as:  

[tex]z=\frac{X-\mu}{\sigma}[/tex]  

At, the 97th percentile of a normal distribution, the z-score, according to a z-score table, is 1.881.

Therefore, the minimum score, X, needed for a bowler to receive a trophy is:

[tex]1.881=\frac{X-182}{14}\\X=208.334[/tex]

Since only whole point scores are possible, X=209 points.

To find the minimum score at the 97th percentile for bowlers in a league, one must calculate the z-score for the 97th percentile and then apply the formula Score = μ + (z * σ) using the league's mean and standard deviation.

To find the minimum score needed for a bowler to receive a trophy (which is at or above the 97th percentile), we need to use the normal distribution properties. With a mean (μ) of 182 points and a standard deviation (σ) of 14 points, we can find the z-score corresponding to the 97th percentile using a z-table or a calculator with normal distribution functions. Once we have the z-score, we can use the formula:

Score = μ + (z * σ)

to calculate the score that corresponds to the 97th percentile.

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Answer:

10 ft x 10 ft

Area = 100 ft^2

Step-by-step explanation:

Let 'S' be the length of the southern boundary fence and 'W' the length of the eastern and western sides of the fence.

The total area is given by:

[tex]A=S*W[/tex]

The cost function is given by:

[tex]\$ 120 = \$3*2W+\$6*S\\20 = W+S\\W = 20-S[/tex]

Replacing that relationship into the Area function and finding its derivate, we can find the value of 'S' for which the area is maximized when the derivate equals zero:

[tex]A=S*(20-S)\\A=20S-S^2\\\frac{dA}{dS} = \frac{d(20S-S^2)}{dS}\\0= 20-2S\\S=10[/tex]

If S=10 then W =20 -10 = 10

Therefore, the largest area enclosed by the fence is:

[tex]A=S*W\\A=10*10 = 100\ ft^2[/tex]

Parameterize the line segments by

[tex]\vec r(t)=(1-t)\langle1,1,1\rangle+t\langle2,2,2\rangle=\langle1+t,1+t,1+t\rangle[/tex]

and

[tex]\vec s(t)=(1-t)\langle2,2,2\rangle+t\langle3,6,8\rangle=\langle2+t,2+4t,2+6t\rangle[/tex]

both with [tex]0\le t\le1[/tex]. Then

[tex]\vec r'(t)=\langle1,1,1\rangle[/tex]

[tex]\vec s'(t)=\langle1,4,6\rangle[/tex]

so that the work done by [tex]\vec F(x,y,z)=\langle\sin(x+y),xy,x^2z\rangle[/tex] over the respective line segments is given by

[tex]\displaystyle\int_{C_1}\vec F\cdot\mathrm d\vec r=\int_0^1\langle\sin(2+2t),(1+t)^2,(1+t)^3\rangle\cdot\langle1,1,1\rangle\,\mathrm dt[/tex]

[tex]=\displaystyle\int_0^1\sin(2+2t)+(1+t)^2+(1+t)^3\,\mathrm dt=\boxed{\frac{73+6\cos2-6\cos4}{12}}[/tex]

and

[tex]\displaystyle\int_{C_2}\vec F\cdot\mathrm d\vec s=\int_0^1\langle\sin(4+5t),(2+t)(2+4t),(2+t)^2(2+6t)\rangle\cdot\langle1,4,6\rangle\,\mathrm dt[/tex]

[tex]=\displaystyle\int_0^1\sin(4+5t)+4(2+t)(2+4t)+6(2+t)^2(2+6t)\,\mathrm dt=\boxed{\frac{3695+3\cos4-3\cos9}{15}}[/tex]

(both measured in Newton-meters)

To compute the work done by the force in this scenario, we need to apply the principle of line integrals of force with respect to displacement, calculating work for each segment of the trajectory separately and then adding these results.

Computing work done by the given force while moving an object generally involves utilizing the line integral of the force with respect to displacement. The principle is simplistically shown in the equation W = F⋅d = Fd cos θ, where W represents work, F denotes force, and d symbolizes displacement, with '⋅' denoting the dot product, and cos θ representing the cosine of the angle between the force and displacement vectors. F can break down into its components, such as Fx, Fy, Fz, and similarly, d can be broken down to dx, dy, dz. Using these, we can express work for three dimensions as dW = Fxdx + Fydy + Fzdz which extends the notion of work done to three-dimensional space. This concept is categorically illustrated while calculating the infinitesimal work done by a variable force.

The trajectory (path) for this problem can be divided into two line segments, one from (1, 1, 1) to (2, 2, 2) and the other from (2, 2, 2) to (3, 6, 8). The work for each segment can be calculated independently, based on the variable force function and the displacement during each segment, after which the two results can be added up to determine the total work done.

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[tex]\vec F(x,y,z)=y\,\vec\imath+x\,\vec\jmath+3\,\vec k[/tex]

is conservative if there is a scalar function [tex]f(x,y,z)[/tex] such that [tex]\nabla f=\vec F[/tex]. This would require

[tex]\dfrac{\partial f}{\partial x}=y[/tex]

[tex]\dfrac{\partial f}{\partial y}=x[/tex]

[tex]\dfrac{\partial f}{\partial z}=3[/tex]

(or perhaps the last partial derivative should be 4 to match up with the integral?)

From these equations we find

[tex]f(x,y,z)=xy+g(y,z)[/tex]

[tex]\dfrac{\partial f}{\partial y}=x=x+\dfrac{\partial g}{\partial y}\implies\dfrac{\partial g}{\partial y}=0\implies g(y,z)=h(z)[/tex]

[tex]f(x,y,z)=xy+h(z)[/tex]

[tex]\dfrac{\partial f}{\partial z}=3=\dfrac{\mathrm dh}{\mathrm dz}\implies h(z)=3z+C[/tex]

[tex]f(x,y,z)=xy+3z+C[/tex]

so [tex]\vec F[/tex] is indeed conservative, and the gradient theorem (a.k.a. fundamental theorem of calculus for line integrals) applies. The value of the line integral depends only the endpoints:

[tex]\displaystyle\int_{(1,2,3)}^{(5,7,-2)}y\,\mathrm dx+x\,\mathrm dy+3\,\mathrm dz=\int_{(1,2,3)}^{(5,7,-2)}\nabla f(x,y,z)\cdot\mathrm d\vec r[/tex]

[tex]=f(5,7,-2)-f(1,2,3)=\boxed{18}[/tex]

Answer:

On this case the 0.09 value is not included on the interval so we can say that the statement on the television news broadcast reports, is a value away from the real proportion at least at 95% of confidence.

Step-by-step explanation:

1) Notation and definitions

[tex]X=34[/tex] number of people living at USA with a particular disease.

[tex]n=527[/tex] random sample taken

[tex]\hat p=\frac{34}{527}=0.0645[/tex] estimated proportion of people living at USA with a particular disease

[tex]p[/tex] true population proportion of people living at USA with a particular disease.

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".  

The population proportion have the following distribution

[tex]p \sim N(p,\sqrt{\frac{\hat p(1-\hat p)}{n}})[/tex]

In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 95% of confidence, our significance level would be given by [tex]\alpha=1-0.95=0.05[/tex] and [tex]\alpha/2 =0.025[/tex]. And the critical value would be given by:

[tex]z_{\alpha/2}=-1.96, z_{1-\alpha/2}=1.96[/tex]

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

[tex]\hat p \pm z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}[/tex]

If we replace the values obtained we got:

[tex]0.0645 - 1.96\sqrt{\frac{0.0645(1-0.0645)}{527}}=0.0435[/tex]

[tex]0.0645 + 1.96\sqrt{\frac{0.0645(1-0.0645)}{527}}=0.0855[/tex]

The 95% confidence interval would be given by (0.0435;0.0855)

On this case the 0.09 value is not included on the interval so we can say that the statement on the television news broadcast reports, is a value away from the real mean at least at 95% of confidence.

Answer:

[tex]P(X=k) = \displaystyle\frac{(2.5)^ke^{-2.5}}{k!}[/tex]

Step-by-step explanation:

This is a typical example where the Poisson distribution is a good choice to model the situation.

In this case we have an interval of time of 50 milliseconds as average time for the server to address one request and 50 requests per second.  

By cross-multiplying we determine the expected value of requests every 50 milliseconds.  

We know 1 second = 1,000 milliseconds

50 requests __________ 1000 milliseconds

 x requests __________ 50 milliseconds

50/x = 1000/50 ===> x = 2.5  

and the expected value is 2.5 requests per interval of 50 milliseconds.

According to the Poisson distribution, the probability of k events in 50 milliseconds equals

[tex]\bf P(X=k) = \displaystyle\frac{(2.5)^ke^{-2.5}}{k!}[/tex]

Answer:

0.6284,0.4335,0.1953.0.9786

Step-by-step explanation:

Given that X the number of material anomalies occurring in a particular region of an aircraft gas-turbine disk is following a poisson distribution with parameter 4

a) [tex]P(X\leq 4)=0.6284\\P(X<4)=0.4335[/tex]

b) [tex]P(4\leq x<5)\\=P(4)=0.1953\\[/tex]

c) P([tex]5\leq x)[/tex]=0.8046

d) the probability that the number of anomalies does not exceed the mean value by more than one standard deviation

=[tex]P(0\leq X\leq 8)\\=F(8)-F(0)\\=0.9786[/tex]

Answer:

add the decimals thats all

Step-by-step explanation:

P-value for the airline's claim that the no-show rate for passengers booked on its flights is less than 6% is 0.2061. Hence, option (B) is correct.

Given that, an airline claims that the no-show rate for passengers booked on its flights is less than 6%.

From this claim, it is clear that:

Null hypothesis :  Proportion of no-show rate for passengers booked on its flights, P = 6%. i.e. H₀: P = 0.06

Alternative hypothesis: Proportion of no-show rate for passengers booked on its flights P < 6%, i.e. P < 0.06.

Out of 380 randomly selected reservations, 19 were no-shows

Sample proportion, [tex]\hat{p} = \frac{19}{380}[/tex]  = 0.05

Then, the standard error for the sample size of 380 is:

[tex]\text{S.E.} = \sqrt{{\frac{p(1-p)}{n}[/tex]

[tex]\text{S.E.} = \sqrt{{\frac{0.06(1-0.06)}{380}[/tex]

[tex]\text{S.E.} = 0.012[/tex]

Now calculating the test statistic

[tex]z = \frac{( \hat{p} - P)}{S.E}[/tex]

[tex]z = \frac{(0.05 - 0.06)}{0.012 }[/tex]

z = -0.833

p value for z = -0.083 is 0.2061 (From the normal table).

Hence, the p-value for a test of the airline's claim is 0.2061. Option (B) is correct.

Learn more about p-value here:

https://brainly.com/question/33325466

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Final answer:

To find the P-value for the hypothesis test, calculate the sample proportion and standard error, then use the standard normal distribution to calculate the z-score and find the corresponding P-value.

Explanation:

To find the P-value for the hypothesis test, we need to use the given data and perform calculations.

First, we need to calculate the sample proportion, which is the number of no-shows divided by the total number of reservations: 19/380 = 0.05.

Next, we calculate the standard error of the sample proportion using the formula: √((p' * (1 - p')) / n), where p' is the sample proportion and n is the sample size. In this case, the standard error is √((0.05 * (1 - 0.05)) / 380) = 0.014.

Finally, we use the standard normal distribution to calculate the z-score and find the corresponding P-value. In this case, the observed proportion is less than the claimed proportion, so we use a one-tailed test and calculate the z-score as (observed proportion - claimed proportion) / standard error = (0.05 - 0.06) / 0.014 = -0.714. Looking up the P-value for a z-score of -0.714, we find that it is approximately 0.4714.

Therefore, the P-value for the test of the airline's claim is approximately 0.4714.

Answer:

Option b) Increase the sample mean

Step-by-step explanation:

Given that a researcher selects a sample from a population with μ = 30 and uses the sample to evaluate the effect of a treatment. After treatment, the sample has a mean of M = 32 and a variance of s2 = 6.

This is a paired test with test statistic

=mean diff/std error

Mean difference would increase if sample mean increases.

This would increase the test statistic

Or otherwise decrease in variance will increase the test statistic

Or Increase in sample size would also increase test statistic

Of all these the II option is definite in increasing the  likelihood of rejecting the null hypothesis because this would definitely increase the chances of rejecting H0.

Others may also have effect but not as much direct as sample mean difference.

Because variance and sample size have influence only upto square root of the difference.