Ivan started the week on page 35 of his book and read 20 pages each night. What page would Ivan be on if he reads for 8 nights

Answer:

Step-by-step explanation:

Friend sells x cups

You sell 3x cups

Total cups sold:: 4x

Ans: x/4x = 1/4 = 25%

Final answer:

The fraction of cups sold by your friend is x/4x, which gives us 25% of the total sales when converted to a percentage.

Explanation:

To find out what percent of the cups sold were sold by your friend, we can set up a simple ratio.

Let's assume your friend sold x cups.

According to the problem, you sold three times as many cups, so you sold 3x cups.

The total number of cups sold is the sum of the cups you both sold, which is x + 3x = 4x cups.

Now, we want to find out what fraction of the total sales x represents.

Since your friend sold x cups out of the total 4x, the fraction is x/4x.

To convert this fraction to a percentage, we multiply it by 100%.

Therefore, the percentage of cups sold by your friend is (x/4x) × 100% = 25%.

This calculation shows that your friend sold 25% of the cups.

Answer:

Step-by-step explanation:

Recall that we say that d | a if there exists an integer k for which a = dk. So, let d = gcd(a,b) and let x, y be integers. Let t = ax+by.

We know that [tex]d | a, d | b[/tex] so there exists integers k,m such that a = kd and b = md. Then,

[tex] t = ax+by = (kd)x+(md)y = d(kx+my)[/tex]. Recall that since k,  x, m, y are integers, then (kx+my) is also an integer. This proves that d | t.

Answer:no

Step-by-step explanation:

expressions don't have equal sides

so it's an equation

please like and Mark as brainliest

Answer: the z- score when x= 186 is 2.643

the mean is 149

this z score tells you that x= 186 is 2.643 standard dev. to the right of the mean

Answer:

[tex]n=\frac{0.45(1-0.45)}{(\frac{0.05}{1.96})^2}=380.32[/tex]  

And rounded up we have that n=381

Step-by-step explanation:

Previous concepts

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".  

Solution to the problem

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 margin of error for the proportion interval is given by this formula:  

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

And on this case we have that [tex]ME =\pm 0.05[/tex] and we are interested in order to find the value of n, if we solve n from equation (a) we got:  

[tex]n=\frac{\hat p (1-\hat p)}{(\frac{ME}{z})^2}[/tex]   (b)  

And replacing into equation (b) the values from part a we got:

[tex]n=\frac{0.45(1-0.45)}{(\frac{0.05}{1.96})^2}=380.32[/tex]  

And rounded up we have that n=381

Final answer:

To estimate the proportion of out-of-state hotel guests with a 95% confidence level and a margin of error of 5%, a sample size of at least 385 guests is needed.

Explanation:

To estimate the proportion of out-of-state guests at a local hotel with a margin of error of no more than 5% and a 95% level of confidence, we can use the formula for determining sample size for a proportion:

n = (Z^2 * p * (1 - p)) / E^2

Where:
- n is the sample size
- Z is the Z-score corresponding to the confidence level (1.96 for 95% confidence)
- p is the preliminary estimate of the proportion (0.45, or 45%, in this case)
- E is the desired margin of error (0.05, or 5%, here)

Substituting the known values, we get:

n = (1.96^2 * 0.45 * (1 - 0.45)) / 0.05^2

n = 384.16

Since we cannot have a fraction of a person, we would round up to the nearest whole number, which gives us a sample size of 385. Therefore, the hotel should sample at least 385 guests to meet their requirements.

Answer:

:=?¿ZV16

Step-by-step explanation:

Answer: [tex]A = 100[/tex] square unit

Step-by-step explanation:

The formula for calculating the area of a square when the diagonal is given is :

[tex]Area = d^{2} / 2[/tex]

That is :

[tex]A = \frac{(10\sqrt{2})^{2}}{2}[/tex]

[tex]A = \frac{100(2)}{2}[/tex]

[tex]A = 100[/tex] square unit

Answer:

radius-4

Step-by-step explanation:

the radius is half of the diameter

Answer:

since the rate at  which concentration inside the cell is proportional to the difference in the concentration of the solute in the blood stream and the concentration within the cell, then the rate of change of concentration within the cell is equals to K(L-C).

Thus, the  required differential equation is Δc/Δt = K( L - C ).

Step-by-step explanation:

The differential equation that models solute diffusion in osmosis is dC/dt = k*(L - C), representing the idea that the rate of change of the solute's concentration within the cell is proportional to the difference between the external and internal concentrations.

The differential equation that models this situation based on osmosis is expressed as dC/dt = k*(L - C), where C is the concentration of a solute within the cell, L is the constant level concentration in the bloodstream, and k is the constant of proportionality.

This equation is a simple expression of the rate of change being proportional to the difference between the outside and inside concentration levels. In this case, the rate of change of the concentration of solutes, dC/dt, is proportional to the difference between the concentration outside the cell, L, and the concentration inside the cell, C. The constant of proportionality is k.

This differential equation is a first order linear differential equation and represents processes that occur widely in nature, including in the diffusion of solutes through membranes in osmosis.

Learn more about Differential Equation here:

https://brainly.com/question/28052300

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Answer:

Probability that this project is going to take more than 18 weeks = 0.99991

Step-by-step explanation:

When independent distributions are combined, the combined mean and combined variance are given through the relation

Combined mean = Σ λᵢμᵢ

(summing all of the distributions in the manner that they are combined)

Combined variance = Σ λᵢ²σᵢ²

(summing all of the distributions in the manner that they are combined)

For this distribution, the total time the project will take = A + B

A ~ (15, 8)

B ~ (16, 4)

Combined mean = μ₁ + μ₂ = 15 + 16 = 31

Combined variance = 1²σ₁² + 1²σ₂² = 8 + 4 = 12

Combined Standard Deviation = √(12) = 3.464 weeks

So, with the right assumption that this combined distribution is a normal distribution

Probability that this project is going to take more than 18 weeks

P(x > 18)

We first normalize/standardize 18

The standardized score for any value is the value minus the mean then divided by the standard deviation.

z = (x - μ)/σ = (18 - 31)/3.464 = - 3.75

The required probability

P(x > 18) = P(z > -3.75)

We'll use data from the normal probability table for these probabilities

P(x > 18) = P(z > -3.75) = 1 - P(x ≤ -3.75)

= 1 - 0.00009 = 0.99991

Hope this Helps!!!

a)

[tex]2a+(3a-7)=2a+3a-7=5a-7[/tex]

b)

[tex](2x-3)+x=2x-3+x=3x-3[/tex]

c)

[tex]7-(5x+4)=7-5x-4=3-5x[/tex]

d)

[tex]3x-(y-2x)=3x-y+2x=5x-y[/tex]

e)

[tex]-(3a+6)-4=-3a-6-4=-3a-10[/tex]

f)

[tex]-(x-2y)-3x=-x+2y-3x=2y-4x[/tex]

g)

[tex](2p-1)+(3+5p)=2p-1+3+5p=7p+2[/tex]

h)

[tex](4-2x)-(7x-2)=4-2x-7x+2=6-9x[/tex]

i)

[tex]-(2x-1)+(3x+1)=-2x+1+3x+1=x+2[/tex]

j)

[tex](2m+4n)+(m-0,5n)=2m+4n+m-0,5n=3m+3,5n[/tex]

k)

[tex](4a-7b)-(2a+3b)=4a-7b-2a-3b=2a-10b[/tex]

l)

[tex]-(2p-r)-(2r-p)=-2p+r-2r+p=-p-r[/tex]

Answer:

[tex]\\\sum_{k=1}^{n+1}\frac{1}{k(k+1)}\\ \\ \\=\sum_{k=1}^n\frac{1}{k(k+1)}+\frac{1}{(n+1)(n+2)} \\ \\ =1-\frac{1}{n+1}+\frac{1}{(n+1)(n+2)}\\ \\ \\ =1+\frac{1-(n+2)}{(n+1)(n+2)} \\ \\ \\ \\\sum_{k=1}^{n+1}\frac{1}{k(k+1)} =1-\frac{1}{n+2}[/tex]

Step-by-step explanation:

The question says; Proof that :

[tex]Let : P\left(n\right)=\sum_{k=1}^n\frac{1}{k(k+1)}=1-\frac{1}{n+1}[/tex]

Base case: P(1) = 1/2.

Inductive step: suppose P(n) has already been proven for some arbitrary n.  The statement P(n+1) is :

[tex]P\left(n+1\right)=\sum_{k=1}^{n+1}\frac{1}{k\left(k+1\right)}=1-\frac{1}{n+2}[/tex]

This concludes the proof by induction.

We Proof that:

The proof abuses the notation P(n) to make reference to the common values of the two sides of the equation to be proved. Moreover, it doesn't makes any sense to define P(n) as the common value of the two sides because it assumes the conclusion that the two sides are equal.

At the very least, the definition of P(n) in the first statement suppose to have be in quote or in parenthesis as shown below.

[tex]Let : P\left(n\right)=(\sum_{k=1}^n\frac{1}{k(k+1)}=1-\frac{1}{n+1})[/tex]

However , P(n) is a statement.

The proof writer confused stating P(n+1) with showing that it must be true; given that P(n) is true.

As such ; the correct proof for P(n+1) is:

[tex]\\\sum_{k=1}^{n+1}\frac{1}{k(k+1)}\\ \\ \\=\sum_{k=1}^n\frac{1}{k(k+1)}+\frac{1}{(n+1)(n+2)} \\ \\ =1-\frac{1}{n+1}+\frac{1}{(n+1)(n+2)}\\ \\ \\ =1+\frac{1-(n+2)}{(n+1)(n+2)} \\ \\ \\ \\\sum_{k=1}^{n+1}\frac{1}{k(k+1)} =1-\frac{1}{n+2}[/tex]

Answer:

add -x to 6x to get 5x then divide by 5 to get 2 x is 2

Step-by-step explanation:hope this helps god bless

Step-by-step explanation:

I think you have to find value of x

-x + 6x = 10

5x = 10

x = 10 / 5 = 2

x = 2

Answer:

Probability is the chance of you getting something

for example the chances (or probability) of the dice landing on 2 or 3 is 3 out of 6 (3/6)

hope this helps!

Answer:

0.22

Step-by-step explanation:

Sample given is 500, so use z-score for the critical value

Given 95% confidence interval;

∝=100-95 =5% =0.05

∝/2 = 0.05/2 =0.025 ----because you are interested with one tail area

1-0.025= 0.975 -----area to the left

proceed to z-table to read 0.975 = 1.96 as the critical value

standard deviation from the question is 2.5 but because this is a sample then;

standard error for the mean is= standard deviation/√sample= 2.5/√500

=0.1118

Margin of error=1.96*0.1118 =0.2191

The margin of error is approximately 0.219 inches. Therefore, the correct answer is approximately 0.22 inches. Option (c) is correct.

To find the margin of error for a 95% confidence interval, we first need to determine the critical value for a 95% confidence interval.

For a normal distribution, with a 95% confidence level, the critical value is approximately 1.96.

Then, we use the formula for the margin of error:

[tex]\[ \text{Margin of Error} = \text{Critical Value} \times \frac{\text{Standard Deviation}}{\sqrt{\text{Sample Size}}} \][/tex]

Given:

Let's calculate the margin of error:

[tex]\[ \text{Margin of Error} = 1.96 \times \frac{2.5}{\sqrt{500}} \][/tex]

[tex]\[ \text{Margin of Error} = 1.96 \times \frac{2.5}{\sqrt{500}} \][/tex]

[tex]\[ \text{Margin of Error} \approx1.96 \times \frac{2.5}{\sqrt{500}} \][/tex]

[tex]\[ \text{Margin of Error} \approx 1.96 \times \frac{2.5}{22.36} \][/tex]

[tex]\[ \text{Margin of Error} \approx 1.96 \times 0.1118 \][/tex]

[tex]\[ \text{Margin of Error} \approx 0.219 \][/tex]

So, the margin of error is approximately 0.219 inches. Therefore, the correct answer is approximately 0.22 inches.

Answer:

The 99% confidence interval for the average fluid content of a can is between 11.54 and 12.66 fluid ounces.

Step-by-step explanation:

We are in posession of the sample's standard deviation, so we use the student t-distribution to solve this question.

The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So

df = 25 - 1 = 24

99% confidence interval

Now, we have to find a value of T, which is found looking at the t table, with 24 degrees of freedom(y-axis) and a confidence level of [tex]1 - \frac{1 - 0.99}{2} = 0.995([tex]t_{995}[/tex]). So we have T = 2.797

The margin of error is:

M = T*s = 2.797*0.2 = 0.56

In which s is the standard deviation of the sample.

The lower end of the interval is the sample mean subtracted by M. So it is 12.1 - 0.56 = 11.54 fluid ounces.

The upper end of the interval is the sample mean added to M. So it is 12.1 + 0.56 = 12.66 fluid ounces.

The 99% confidence interval for the average fluid content of a can is between 11.54 and 12.66 fluid ounces.

Answer:

B

Step-by-step explanation:

Null hypothesis is not rejected if the test statistics show that the difference between observed and expected value is not significant and any difference is only due to chance.

So there is sufficient evidence from statistics that null hypothesis is true and the standard deviation in the pressure required to open a certain valve has not changed.

since evidence has to be there when null hypothesis is tested, optionC,  D and E are rejected.

There is no change in standard deviation so option A and F are rejected.

Answer:

a = 1

Step-by-step explanation:

x = 8

3x - y = 9

3(8) - y = 9

24 - y = 9

-y = 9 - 24

-y = -15

y = 15

ax + y = 23

a(8) + 15 = 23

8a = 23 - 15

8a = 8

a = 8/8

a = 1

Answer:

-8/45

I hope this helped!

Step-by-step explanation:

-2/3 ÷ 3 3/4

Answer:

0.266

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

[tex]\mu = 5.85, \sigma = 0.24[/tex]

Find the probability that a randomly selected mandarin orange from this farm has a diameter larger than 6.0 cm.

This is 1 subtracted by the pvalue of Z when X = 6.

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]Z = \frac{6 - 5.85}{0.24}[/tex]

[tex]Z = 0.625[/tex]

[tex]Z = 0.625[/tex] has a pvalue of 0.734

1 - 0.734 = 0.266

Answer:

[tex]P(X>6)=P(\frac{X-\mu}{\sigma}>\frac{6-\mu}{\sigma})=P(Z>\frac{6-5.85}{0.24})=P(z>0.625)[/tex]

And we can find this probability using the complement rule and the normal standard table or excel:

[tex]P(z>0.625)=1-P(z<0.625)=1-0.734= 0.266[/tex]

Step-by-step explanation:

Previous concepts

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 Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

Solution to the problem

Let X the random variable that represent the diameters of mandarin oranges of a population, and for this case we know the distribution for X is given by:

[tex]X \sim N(5.85,0.24)[/tex]  

Where [tex]\mu=5.85[/tex] and [tex]\sigma=0.24[/tex]

We are interested on this probability

[tex]P(X>6)[/tex]

And the best way to solve this problem is using the normal standard distribution and the z score given by:

[tex]z=\frac{x-\mu}{\sigma}[/tex]

If we apply this formula to our probability we got this:

[tex]P(X>6)=P(\frac{X-\mu}{\sigma}>\frac{6-\mu}{\sigma})=P(Z>\frac{6-5.85}{0.24})=P(z>0.625)[/tex]

And we can find this probability using the complement rule and the normal standard table or excel:

[tex]P(z>0.625)=1-P(z<0.625)=1-0.734= 0.266[/tex]

Answer:

(a) The estimated proportion of defective in the population is 0.05.

(b) The 95% confidence interval for the proportion defective cups is (2.5%, 7.5%).

(c) Zack does not needs to return the lots.

Step-by-step explanation:

Let X = number of defective cups.

The random sample of cups selected is of size, n = 300.

The number of defective cps in the sample is, X = 15.

(a)

The proportion of the defective cups in the population can be estimated by the sample proportion because the sample selected is quite large.

The sample proportion of defective cups is:

[tex]\hat p=\frac{X}{n}=\frac{15}{300}=0.05[/tex]

Thus, the estimated proportion of defective in the population is 0.05.

(b)

The (1 - α)% confidence interval for population proportion is:

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

Compute the critical value of z for 95% confidence level as follows:

[tex]z_{\alpha/2}=z_{0.05/2}=z_{0.025}=1.96[/tex]

Compute the 95% confidence interval for p as follows:

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

     [tex]=0.05 \pm 1.96\times\sqrt{\frac{0.05(1-0.05)}{300}}[/tex]

     [tex]=0.05\pm 0.025\\=(0.025, 0.075)\\[/tex]

Thus, the 95% confidence interval for the proportion defective cups is (2.5%, 7.5%).

(c)

It is provided that Zack has an agreement with his supplier that he is to return lots that are 10% or more defective.

The 95% confidence interval for the proportion defective is (2.5%, 7.5%). This implies that 95% of the lots have 2.5% to 7.5% defective items.

Thus, Zack does not needs to return the lots.

The estimated proportion of defective in the population is 0.05 and the 95 percent confidence interval for the proportion defective is (0.025,0.075).

Given :

a) The formula given below is used in order to determine the estimated proportion of defective in the population.

[tex]\hat{p} = \dfrac{X}{n}[/tex]

[tex]\hat{p} = \dfrac{15}{300}[/tex]

[tex]\hat{p} = 0.05[/tex]

So, the estimated proportion of defective in the population is 0.05.

b) The below formula is used in order to determine the 95 percent confidence interval for the proportion defective.

[tex]CI =\hat{p}\pm z_{\alpha /2}\times \sqrt{ \dfrac{\hat{p}(1-\hat{p})}{n}}[/tex]

Now, substitute the known terms in the above expression.

[tex]CI =0.05 \pm 1.96\times \sqrt{ \dfrac{0.05(1-0.05))}{300}}[/tex]

[tex]CI = 0.05\pm 0.025[/tex]

So, the 95 percent confidence interval for the proportion defective is (0.025,0.075).

c) According to the given data, Zack has an agreement with his supplier that he is to return lots that are 10 percent or more defective.

So, from the above calculation, it can be concluded that he did not have to return the lots.

For more information, refer to the link given below:

https://brainly.com/question/10951564

Answer:

Step-by-step explanation:

Hello!

The mean life of 16 lithium batteries was estimated with a 95% CI:

(628.5, 661.5) hours

Assuming that the variable "X: Duration time (life) of a lithium battery(hours)" has a normal distribution and the statistic used to estimate the population mean was s Student's t, the formula for the interval is:

[X[bar]±[tex]t_{n-1;1-\alpha /2}* \frac{S}{\sqrt{n} }[/tex]]

The amplitude of the interval is calculated as:

a= Upper bond - Lower bond

a= [X[bar]+[tex]t_{n-1;1-\alpha /2}* \frac{S}{\sqrt{n} }[/tex]] -[X[bar]-[tex]t_{n-1;1-\alpha /2}* \frac{S}{\sqrt{n} }[/tex]]

and the semiamplitude (d) is half the amplitude

d=(Upper bond - Lower bond)/2

d=([X[bar]+[tex]t_{n-1;1-\alpha /2}* \frac{S}{\sqrt{n} }[/tex]] -[X[bar]-[tex]t_{n-1;1-\alpha /2}* \frac{S}{\sqrt{n} }[/tex]] )/2

d= [tex]t_{n-1;1-\alpha /2}* \frac{S}{\sqrt{n} }[/tex]

The sample mean marks where the center of the calculated interval will be. The terms of the formula that affect the width or amplitude of the interval is the value of the statistic, the sample standard deviation and the sample size.

Using the semiamplitude of the interval I'll analyze each one of the posibilities to see wich one will result in an increase of its amplitude.

Original interval:

Amplitude: a= 661.5 - 628.5= 33

semiamplitude d=a/2= 33/2= 16.5

1) Having a sample with a larger standard deviation.

The standard deviation has a direct relationship with the semiamplitude of the interval, if you increase the standard deviation, it will increase the semiamplitude of the CI

↑d= [tex]t_{n-1;1-\alpha /2}[/tex] * ↑S/√n

2) Using a 99% confidence level instead of 95%.

d= [tex]t_{n_1;1-\alpha /2}[/tex] * S/√n

Increasing the confidence level increases the value of t you will use for the interval and therefore increases the semiamplitude:

95% ⇒ [tex]t_{15;0.975}= 2.131[/tex]

99% ⇒ [tex]t_{15;0.995}= 2.947[/tex]

The confidence level and the semiamplitude have a direct relationship:

↑d= ↑[tex]t_{n_1;1-\alpha /2}[/tex] * S/√n

3) Removing an outlier from the data.

Removing one outlier has two different effects:

1) the sample size is reduced in one (from 16 batteries to 15 batteries)

2) especially if the outlier is far away from the rest of the sample, the standard deviation will decrease when you take it out.

In this particular case, the modification of the standard deviation will have a higher impact in the semiamplitude of the interval than the modification of the sample size (just one unit change is negligible)

↓d= [tex]t_{n_1;1-\alpha /2}[/tex] * ↓S/√n

Since the standard deviation and the semiamplitude have a direct relationship, decreasing S will cause d to decrease.

4) Using a 90% confidence level instead of 95%.

↓d= ↓[tex]t_{n_1;1-\alpha /2}[/tex] * S/√n

Using a lower confidence level will decrease the value of t used to calculate the interval and thus decrease the semiamplitude.

5) Testing 10 batteries instead of 16. and 6) Testing 24 batteries instead of 16.

The sample size has an indirect relationship with the semiamplitude if the interval, meaning that if you increase n, the semiamplitude will decrease but if you decrease n then the semiamplitude will increase:

From 16 batteries to 10 batteries: ↑d= [tex]t_{n_1;1-\alpha /2}[/tex] * S/√↓n

From 16 batteries to 24 batteries: ↓d= [tex]t_{n_1;1-\alpha /2}[/tex] * S/√↑n

I hope this helps!

Answer:

A) σ_x' = 1.4142

B) σ_x' = 1.4135

C) σ_x' = 1.4073

D) σ_x' = 1.343

Step-by-step explanation:

We are given;

σ = 10

n = 50

A) when size is infinite, the standard deviation of the sample mean is given by the formula;

σ_x' = σ/√n

Thus,

σ_x' = 10/√50

σ_x' = 1.4142

B) size is given, thus, the standard deviation of the sample mean is given by the formula;

σ_x' = (σ/√n)√((N - n)/(N - 1))

Thus, with size of N = 50,000, we have;

σ_x' = 1.4142 x √((50000 - 50)/(50000 - 1))

σ_x' = 1.4142 x 0.9995

σ_x' = 1.4135

C) at N = 5000;

σ_x' = 1.4142 x √((5000 - 50)/(5000 - 1))

σ_x' = 1.4073

D) at N = 500;

σ_x' = 1.4142 x √((500 - 50)/(500 - 1))

σ_x' = 1.343

The value of the standard error in each of the following are:

A) σ_x' = 1.4142

B) σ_x' = 1.4135

C) σ_x' = 1.4073

D) σ_x' = 1.343

Given:

σ = 10

n = 50

A) when size is infinite, the standard deviation of the sample mean is given by the formula;

σ_x' = σ/√n

Thus,

σ_x' = 10/√50

σ_x' = 1.4142

B) size is given, thus, the standard deviation of the sample mean is given by the formula;

σ_x' = (σ/√n)√((N - n)/(N - 1))

Thus, with size of N = 50,000, we have;

σ_x' = 1.4142 x √((50000 - 50)/(50000 - 1))

σ_x' = 1.4142 x 0.9995

σ_x' = 1.4135

C) at N = 5000;

σ_x' = 1.4142 x √((5000 - 50)/(5000 - 1))

σ_x' = 1.4073

D) at N = 500;

σ_x' = 1.4142 x √((500 - 50)/(500 - 1))

σ_x' = 1.343

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Answer:

2z^2+7z+6

because you multiply 2z with z and 2 then multiply 3 with z and 2 and combine like terms

Answer:

(2z + 3) (z +2)

combine like terms: (2z⋅z)+(2z⋅2)+3z+(3⋅2)

2z^2+7z+6

The correlation coefficient for the data is 0.0091.

This can be obtained by using the formula of correlation coefficient.

The following information is obtained from the table,

∑x  = 10

∑y = 10

∑xy = 42

∑x² = 42

∑y² = 42

The formula for finding the correlation coefficient,

r = n∑xy- ∑x∑y/(n∑x²-(∑x)²)(n∑y²-(∑y)²)

r = (5×42)-(10×10)/((5×42)-10²)((5×42)-10²)

 =[tex]\frac{110}{110.110}[/tex]

=[tex]\frac{1}{110}[/tex]

=0.0091

Thus, the correlation coefficient of the given data is 0.0091.

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Answer:

B) 1

Step-by-step explanation:

Got it right 2024 edg

Answer:

77.34% of students will score below 555 in a typical year

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

SAT:

[tex]\mu = 480, \sigma = 100[/tex]

(a) An engineering school sets 555 as the minimum SAT math score for new students. What percentage of students will score below 555 in a typical year?

This is the pvalue of Z when X = 555. So

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]Z = \frac{555 - 480}{100}[/tex]

[tex]Z = 0.75[/tex]

[tex]Z = 0.75[/tex] has a pvalue of 0.7734.

77.34% of students will score below 555 in a typical year

Answer:

There is statistical evidence to support the claim that the goal of reducing the rate of undercutting to less than 5% has been met.

P-value=0.01923.

Step-by-step explanation:

We have to test the hypothesis that the proportion of defective circuits is under 5%.

Then, the null and alternative hypothesis are:

[tex]H_0: \pi=0.05\\\\H_a:\pi<0.05[/tex]

We will assume a level of significance of 0.05.

The sample, of size n=1000, has 35 defecteive circuits, so the sample proportion is:

[tex]p=35/1000=0.035[/tex]

The standard error is calculated as if the null hypothesis is true, so it is:

[tex]\sigma_p=\sqrt{\dfrac{\pi(1-\pi)}{n}}=\sqrt{\dfrac{0.05*0.95}{1000}}}=\sqrt{0.0000475}=0.007[/tex]

The z-statistic can be calculated as:

[tex]z=\dfrac{p-\pi+0.5/n}{\sigma_p}=\dfrac{0.035-0.050+0.5/1000}{0.007}=\dfrac{-0.0145}{0.007}= -2.07[/tex]

For this one-tailed test, the P-value is:

[tex]P-value=P(z<-2.07)=0.01923[/tex]

As the P-value is smaller than the significance level, the effect is significant and the null hypothesis is rejected. There is statistical evidence to support the claim that the goal of reducing the rate of undercutting to less than 5% has been met.

The question asks for various measures of an exponentially distributed statistic, the distance between flaws on a cable. To find these, we use formulas specific to the exponential distribution; for example, the median is calculated as ln(2) * mean, and the standard deviation is equal to the mean, computed with the formula σ = √variance = √mean.

The problem can be modeled using the exponential distribution in statistics. The exponential distribution is often used to model the time elapsed between events in a Poisson point process, or in our case, the distance between the flaws in the cable.

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In this problem, we're calculating properties of an exponential distribution representing the distances between flaws in a cable. The probability that a given distance is more than 15m is 22.31%, and between 8m and 20m is 48.57%. The median distance is around 8.317m, the standard deviation is 12m, and the 65th percentile is about 14.791m.

In this problem, the distance between flaws on a long cable is exponentially distributed with a mean of 12 m. That means the lambda (λ) parameter's rate of occurring flaws per meter is 1/mean, or approximately 0.0833.

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Answer:

The pairs are matched

Step-by-step explanation:

A. A line through the origin that makes an angle of [tex]\pi/6[/tex] with the positive x-axis.

Given a line through the origin that makes an angle of [tex]\pi/6[/tex] with the positive x-axis. The angle which the line makes with the x-axis is [tex]\pi/6[/tex].

B. A vertical line through the point (3, 3).

If a line passes through the point (3,3), x=3 and y=3. The vertical line through the point (3,3) is x=3

C. Given a circle center (h,k) and a center r, the standard form of the equation of the circle is given as:

[tex](x-h)^2+(y-k)^2=r^2[/tex]

Therefore, for a circle with radius 5 and center (2, 3), the standard form equation is:

D. A circle centered at the origin with radius.

For a circle centered at the origin with radius r=4.

The radius of the circle is 4 units.

Answer:

565.2 in²

Step-by-step explanation:

2pi × r × (r + h)

2 × 3.14 × 5 × (5 + 13)

31.4 × 18

565.2