Please answer fast will mark brainliest

I. NEED HELP ASAP PLZ SOMEONE HELP CAN U ANSWER AT LEAST ONE

bonufazy [111]

I might be wrong but I did used to get 12.5 for the third one

4 0

3 years ago

In San Jose a sample of 73 mail carriers showed that 30 had been bitten by an animal during one week. In San Francisco in a samp

dsp73

Answer:

$(0.411-0.7) - 1.96 \sqrt{\frac{0.411(1-0.411)}{73} +\frac{0.7(1-0.7)}{80}}=-0.4401$

$(0.411-0.7) + 1.96 \sqrt{\frac{0.411(1-0.411)}{73} +\frac{0.7(1-0.7)}{80}}=-0.1380$

We are confident at 95% that the difference between the two proportions is between $-0.4401 \leq p_B -p_A \leq -0.1380$

1. -.4401 ≤ p1 - p2 ≤ -.1380

4. The rate of mail carriers being bitten in San Jose is statistically less than the rate San Francisco at α = 5%

Step-by-step explanation:

In San Jose a sample of 73 mail carriers showed that 30 had been bitten by an animal during one week. In San Francisco in a sample of 80 mail carriers, 56 had received animal bites. Is there a significant difference in the proportions? Use a 0.05. Find the 95% confidence interval for the difference of the two proportions. Sellect all correct statements below based on the data given in this problem.

1. -.4401 ≤ p1 - p2 ≤ -.1380

2. -.4401 ≤ p1 - p2 ≤ .1380

3. The rate of mail carriers being bitten in San Jose is statistically greater than the rate San Francisco at α = 5%

4. The rate of mail carriers being bitten in San Jose is statistically less than the rate San Francisco at α = 5%

5. The rate of mail carriers being bitten in San Jose and San Francisco are statistically equal at α = 5%

Solution to the problem

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".

$p_1$ represent the real population proportion for San Jose

$\hat p_1 =\frac{30}{73}=0.411$ represent the estimated proportion for San Jos

$n_1=73$ is the sample size required for San Jose

$p_2$ represent the real population proportion for San Francisco

$\hat p_2 =\frac{56}{80}=0.7$ represent the estimated proportion for San Francisco

$n_2=80$ is the sample size required for San Francisco

$z$ represent the critical value for the margin of error

The population proportion have the following distribution

$p \sim N(p,\sqrt{\frac{p(1-p)}{n}})$

The confidence interval for the difference of two proportions would be given by this formula

$(\hat p_1 -\hat p_1) \pm z_{\alpha/2} \sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1} +\frac{\hat p_2 (1-\hat p_2)}{n_2}}$

For the 95% confidence interval the value of $\alpha=1-0.95=0.05$ and $\alpha/2=0.025$ , with that value we can find the quantile required for the interval in the normal standard distribution.

$z_{\alpha/2}=1.96$

And replacing into the confidence interval formula we got:

$(0.411-0.7) - 1.96 \sqrt{\frac{0.411(1-0.411)}{73} +\frac{0.7(1-0.7)}{80}}=-0.4401$

$(0.411-0.7) + 1.96 \sqrt{\frac{0.411(1-0.411)}{73} +\frac{0.7(1-0.7)}{80}}=-0.1380$

We are confident at 95% that the difference between the two proportions is between $-0.4401 \leq p_B -p_A \leq -0.1380$

Since the confidence interval contains all negative values we can conclude that the proportion for San Jose is significantly lower than the proportion for San Francisco at 5% level.

Based on this the correct options are:

1. -.4401 ≤ p1 - p2 ≤ -.1380

4. The rate of mail carriers being bitten in San Jose is statistically less than the rate San Francisco at α = 5%

8 0

3 years ago

Marco sorted these solid figures into two groups. What is true about the groups?

Lisa [10]

Answer:

All figures in Group 1 appear to have at

least one square face

Step-by-step explanation:

Let us examine each statement and see whether they are true or not.

The first statement is not true. Only one of the figures (square pyramid) has at least one triangular face. It has 4 triangular faces, actually.

The second statement is not true. Only one of the figures is a rectangular prism. The remaining two are triangular pyramid and triangular prism respectively.

The third statement is not true also. Only 1 figure is a triangular prism.

The fourth statement is correct. In group 1, all the figures has at least one square face.

4 0

3 years ago

A square is put together by two triangles. The hypotenuse is 11 cm and width of 8 cm. find the height of triangle?

Aleks [24]

You can use the pythagorean theorem. Make sure that you put the measurements in the right places though. Since 11 is the measure of the hypotenuse, 11 has to be c. When you plug in the information that you have, it should look like this:
8^2 + b^2 = 11^2

Square 8 and 11.
64 + b^2 = 121

Subtract 64 from 121
121 - 64 = 57

The new equation is:
b^2 = 57

Now take the square root of both sides to get b by itself.
b = about 7.5

Answer: The height of the triangle is 7.5

***** Note: I rounded the answer ********************************************

4 0

3 years ago

Luis purchases a collectible model car from a dealer. The dealer advises him that the car’s value (The car value is $75 now) wil

serious [3.7K]

Judging by the question I generated the equation y=75(1.09^x)
x is the amount of years.
So the equation you should get for 10 years is y=75(1.09^10)
The answer you get should get is $177.55

4 0

3 years ago

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