I need help! I can’t get this one solved correctly

Please solve this <br><br> 4 3/4 * 7 2/17<br><br> A. 28<br><br> B. 32<br><br> C. 35<br><br> D. 40

Varvara68 [4.7K]

The answer is B. 32
i home this helps

5 0

3 years ago

How far is 75% of a 2,000-kilometer trip

zubka84 [21]

Answer: 1500 kilometers

75% of a 2000 - kilometer trip is 1500 kilometers

Step-by-step explanation:

6 0

3 years ago

Daniel has lunch at a restaurant and the cost of his meal is $49.00, because of the service, he wants to leave a 5% tip. what is

yuradex [85]

49*1.05= $51.45

Hope it helped

8 0

3 years ago

A researcher compares two compounds (1 and 2) used in the manufacture of car tires that are designed to reduce braking distances

wariber [46]

Answer:

Step-by-step explanation:

Hello!

The objective of this experiment is to compare two compounds, designed to reduce braking distance, used in tire manufacturing to prove if the braking distance of SUV's equipped with tires made with compound 1 is shorter than the braking distance of SUV's equipped with tires made with compound 2.

So you have 2 independent populations, SUV's equipped with tires made using compound 1 and SUV's equipped with tires made using compound 2.

Two samples of 81 braking tests are made and the braking distance was measured each time, the study variables are determined as:

X₁: Braking distance of an SUV equipped with tires made with compound one.

Its sample mean is X[bar]₁= 69 feet

And the Standard deviation S₁= 10.4 feet

X₂: Braking distance of an SUV equipped with tires made with compound two.

Its sample mean is X[bar]₂= 71 feet

And the Standard deviation S₂= 7.6 feet

We don't have any information on the distribution of the study variables, nor the sample data to test it, but since both sample sizes are large enough n₁ and n₂ ≥ 30 we can apply the central limit theorem and approximate the distribution of both variables sample means to normal.

The researcher's hypothesis, as mentioned before, is that the braking distance using compound one is less than the distance obtained using compound 2, symbolically: μ₁ < μ₂

The statistical hypotheses are:

H₀: μ₁ ≥ μ₂

H₁: μ₁ < μ₂

α: 0.05

The statistic to use to compare these two populations is a pooled Z test

$Z= \frac{(X[bar]_1-X[bar]_2)-(Mu_1-Mu_2)}{\sqrt{\frac{S^2_1}{n_1} +\frac{S^2_2}{n_2} } }$

Z ≈ N(0;1)

$Z_{H_0}= \frac{69-71-0}{\sqrt{\frac{108.16}{81} +\frac{57.76}{81} } }= -1.397$

The rejection region if this hypothesis test is one-tailed to the right, so you'll reject the null hypothesis to small values of the statistic. The critical value for this test is:

$Z_{\alpha } = Z_{0.05}= -1.648$

Decision rule:

If $Z_{H_0}$ > -1.648 , then you do not reject the null hypothesis.

If $Z_{H_0}$ ≤ -1.648 , then you reject the null hypothesis.

Since the statistic value is greater than the critical value, the decision is to not reject the null hypothesis.

At a 5% significance level, you can conclude that the average braking distance of SUV's equipped with tires manufactured used compound 1 is greater than the average braking distance of SUV's equipped with tires manufactured used compound 2.

I hope you have a SUPER day!

4 0

4 years ago

the ratio of number of girls and boys in a class of 30 students is 7:8. if 5 new boys students admit in the class. what will be

Stells [14]

<h3>Given:</h3>

The ratio of number of girls and boys in a class of 30 students is 7:8

The ratio of number of girls and boys if 5 new boys admit in the class.

<h3>Assumption:</h3>

Let the number of students be x.

<h3>Solution:</h3>

According to the question,

7x + 8x = 30

or, 15x = 30

or, x = $\large {\tt {\frac{30}{15}}}$

or, x = 2

7x = 7(2) = 14

8x = 8(2) = 16

There were 14 boys and 16 girls in the school.

After admitting 5 new boys, we get

14 + 5 = <u>19</u>

The ratio of number of girls and boys now is 16:19.

<h2>Answer:</h2>

5 0

3 years ago