Can someone help me with this Edginuity question?

Brody sliced a small pizza into 12 equal pieces. He ate ~ of the pizza. Which model represents three-fourths of twelve?

EastWind [94]

Answer:

Pie chart.

9 slices

Step-by-step explanation:

Three-fourths of twelve is written as

3/4 * 12

i.e 3/4 is multiplied by 12

multiplying 3 by 12 gives 36 and dividing it by 4 gives 9 .

So Brody ate 9 slices out of the 12 slices of pizza.

The graph used in circular shape is called the a pie chart.

If we draw a pie chart 3 of the 4 sections will be colored showing that he ate 3/4 of the pizza representing 9 slices out of 12.

Each slice can be represented separately.

8 0

3 years ago

Having trouble answering these

tensa zangetsu [6.8K]

Answer:

i can't see clearly

Step-by-step explanation:

7 0

3 years ago

A large box contains 10,000 ball bearings. A random sample of 120 is chosen. The sample mean diameter is 10 mm, and the standard

ser-zykov [4K]

Answer:

Option a) A 95% confidence interval for the mean diameter of the 120 bearings in the sample is $10 \pm 1.96\displaystyle\frac{0.24}{\sqrt{120}}$ .

Step-by-step explanation:

We are given the following information in the question:

Sample size, n = 120

Sample mean = 10 mm

Standard Deviation = 0.24 mm

Formula:

$\mu \pm z_{critical}\frac{\sigma}{\sqrt{n}}$

$z_{critical}\text{ at}~\alpha_{0.05} = 1.96$

$10 \pm 1.96\displaystyle\frac{0.24}{\sqrt{120}}$

Hence, the correct interpretation for the confidence interval is given by option a).

A 95% confidence interval for the mean diameter of the 120 bearings in the sample is $10 \pm 1.96\displaystyle\frac{0.24}{\sqrt{120}}$ .

We have to consider the factor of sampling of 120 ball bearings from a population of 10,000 ball bearings.

8 0

3 years ago

25POINTS

svp [43]

Answer:

Its volume for a sphere where radius is 3cm

The real volume is 113.04 cubic meter

7 0

3 years ago

A survey sampled men and women workers and asked if they expected to get a raise or promotion this year. Suppose the survey samp

Oksanka [162]

Answer:

a)

The null hypothesis is: $H_0: p_M - p_W = 0$

The alternative hypothesis is: $H_1: p_M - p_W > 0$

b) For men is of 0.49 and for women is of 0.36.

c) The p-value of the test is 0.0039 < 0.01, which means that the results are statistically significant so that you can conclude a greater proportion of men expect to get a raise or a promotion this year.

Step-by-step explanation:

Before solving this question, we need to understand the central limit theorem and subtraction of normal variables.

Central Limit Theorem

The Central Limit Theorem establishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

For a proportion p in a sample of size n, the sampling distribution of the sample proportion will be approximately normal with mean $\mu = p$ and standard deviation $s = \sqrt{\frac{p(1-p)}{n}}$

Subtraction between normal variables:

When two normal variables are subtracted, the mean is the difference of the means, while the standard deviation is the square root of the sum of the variances.

Men:

98 out of 200, so:

$p_M = \frac{98}{200} = 0.49$

$s_M = \sqrt{\frac{0.49*0.51}{200}} = 0.0353$

Women:

72 out of 200, so:

$p_W = \frac{72}{200} = 0.36$

$s_W = \sqrt{\frac{0.36*0.64}{200}} = 0.0339$

a. State the hypothesis test in terms of the population proportion of men and the population proportion of women.

At the null hypothesis, we test if the proportion are similar, that is, if the subtraction of the proportions is 0, so:

$H_0: p_M - p_W = 0$

At the alternative hypothesis, we test if the proportion of men is greater, that is, the subtraction is greater than 0, so:

$H_1: p_M - p_W > 0$

b. What is the sample proportion for men? For women?

For men is of 0.49 and for women is of 0.36.

c. Use α= 0.01 level of significance. What is the p-value and what is your conclusion?

From the sample, we have that:

$X = p_M - p_W = 0.49 - 0.36 = 0.13$

$s = \sqrt{s_M^2+s_W^2} = \sqrt{0.0353^2 + 0.0339^2} = 0.0489$

The test statistic is:

$z = \frac{X - \mu}{s}$

In which X is the sample mean, $\mu$ is the value tested at the null hypothesis, and s is the standard error, so:

$z = \frac{0.13 - 0}{0.0489}$

$z = 2.66$

P-value of the test and decision:

The p-value of the test is the probability of finding a difference above 0.13, which is the p-value of z = 2.66.

Looking at the z-table, z = 2.66 has a p-value of 0.9961.

1 - 0.9961 = 0.0039.

The p-value of the test is 0.0039 < 0.01, which means that the results are statistically significant so that you can conclude a greater proportion of men expect to get a raise or a promotion this year.

4 0

3 years ago