Help this is trigonometry I think 35 points!!

A math teacher was frustrated at the number of students leaving their graphing calculator behind in her classroom at the end of

AleksAgata [21]

Answer:

The 99% confidence interval for the proportion of all students at this school who have their names written on their graphing calculators is (0.4652, 0.8148).

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of $\pi$ , and a confidence level of $1-\alpha$ , we have the following confidence interval of proportions.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

z is the zscore that has a pvalue of $1 - \frac{\alpha}{2}$ .

A random sample of 50 students is selected, and of the students questioned, 32 had their names written on their graphing calculators.

This means that $n = 50, \pi = \frac{32}{50} = 0.64$

99% confidence level

So $\alpha = 0.01$ , z is the value of Z that has a pvalue of $1 - \frac{0.01}{2} = 0.995$ , so $Z = 2.575$ .

The lower limit of this interval is:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.64 - 2.575\sqrt{\frac{0.64*0.36}{50}} = 0.4652$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.64 + 2.575\sqrt{\frac{0.64*0.36}{50}} = 0.8148$

The 99% confidence interval for the proportion of all students at this school who have their names written on their graphing calculators is (0.4652, 0.8148).

8 0

3 years ago

Find the area of the figure and round to the nearest hundredth

sweet [91]

Answer:

285.45

Step-by-step explanation:

If you multiply 17.3x16.5, you get 285.45, and when you round to the nearest hundredth, well, it gives you the same answer.

Therefore, the answer is 285.45.

5 0

3 years ago

Suppose that a point moves along some unknown curve y = f(x) in such a way that, at

spayn [35]

Answer:

f(x) = x³/3 - 5/3

Step-by-step explanation:

To get the equation of the curve, integrate the slope

f(x) = integral of x²

= (x^(2+1))/3 + c

f(x) = x³/3 + c

To find c, use the given point

1 = 2³/3 + c

c = 1 - 8/3 = -5/3

f(x) = x³/3 - 5/3

8 0

3 years ago

I don’t know what to do

solong [7]

Answer:

Step-by-step explanation:

slope is what is infront of x and y intercept is the number after the slope

6 0

4 years ago

Read 2 more answers

Consider a chemical company that wishes to determine whether a new catalyst, catalyst XA-100, changes the mean hourly yield of i

kolezko [41]

Answer:

Null hypothesis: $\mu = 750$

Alternative hypothesis: $\mu \neq 750$

$t=\frac{811-750}{\frac{19.647}{\sqrt{5}}}=6.943$

$p_v =2*P(t_{4}>6.943)=0.00226$

If we compare the p value and a significance level assumed $\alpha=0.05$ we see that $p_v$ so we can conclude that we reject the null hypothesis, and the actual true mean is significantly different from 750 pounds per hour.

Step-by-step explanation:

Data given and notation

Data: 801, 814, 784, 836,820

We can calculate the sample mean and sample deviation with the following formulas:

$\bar X =\frac{\sum_{i=1}^n X_i}{n}$

$s=\sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}}$

$\bar X=811$ represent the sample mean

$s=19.647$ represent the standard deviation for the sample

$n=5$ sample size

$\mu_o =750$ represent the value that we want to test

$\alpha$ represent the significance level for the hypothesis test.

t would represent the statistic (variable of interest)

$p_v$ represent the p value for the test (variable of interest)

State the null and alternative hypotheses to be tested

We need to conduct a hypothesis in order to determine if the mean is different from 750 pounds per hour, the system of hypothesis would be:

Null hypothesis: $\mu = 750$

Alternative hypothesis: $\mu \neq 750$

Compute the test statistic

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

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$ (1)

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

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

$t=\frac{811-750}{\frac{19.647}{\sqrt{5}}}=6.943$

Now we need to find the degrees of freedom for the t distirbution given by:

$df=n-1=5-1=4$

What do you conclude?

Compute the p-value

Since is a two tailed test the p value would be:

$p_v =2*P(t_{4}>6.943)=0.00226$

If we compare the p value and a significance level assumed $\alpha=0.05$ we see that $p_v$ so we can conclude that we reject the null hypothesis, and the actual true mean is significantly different from 750 pounds per hour.

4 0

3 years ago