Can someone help me on this problem?

How much of each ingredient will Mrs. Roberts need in order to make 200 servings? Show or explain the steps you used to answer t

vitfil [10]

Pretzels:2×200=400 200×1/2=100 100+400=500cups
Dry Cereal: 4×200=800 200×1/2=100 800+100=900cups
Peanuts:2×200=400 200×1/4=50 400+50=450cups
Raisins:200×3/4=150cups

4 0

3 years ago

An airline's public relations department claims that when luggage is lost, 88% is recovered and delivered to its owner within 24

melisa1 [442]

Answer:

The hypothesis statements are: H0: p = 0.88 versus HA : p < 0.88

The p-value of the test is 0.2483 > 0.1, which means that the data does not provide sufficient evidence to conclude that the proportion of times that luggage is returned within 24 hours is less than 0.88.

Step-by-step explanation:

Test if the proportion of times that luggage is returned within 24 hours is less than 0. 88

At the null hypothesis, we test if the proportion is of 0.88, that is:

$H_0: p = 0.88$

At the alternate hypothesis, we test if this proportion is less than 0.88, that is:

$H_a: p < 0.88$

The hypothesis statements are: H0: p = 0.88 versus HA : p < 0.88

The test statistic is:

$z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}$

In which X is the sample mean, $\mu$ is the value tested at the null hypothesis, $\sigma$ is the standard deviation and n is the size of the sample.

0.88 is tested at the null hypothesis:

This means that $\mu = 0.88, \sigma = \sqrt{0.88*0.12}$

A consumer group who surveyed a large number of air travelers found that 138 out of 160 people who lost luggage on that airline were reunited with the missing items by the next day.

This means that $n = 160, X = \frac{138}{160} = 0.8625$

Test statistic:

$z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}$

$z = \frac{0.8625 - 0.88}{\frac{\sqrt{0.88*0.12}}{\sqrt{160}}}$

$z = -0.68$

P-value of the test:

The p-value of the test is the probability of finding a sample proportion below 0.8625, which is the p-value of z = -0.68.

Looking at the z-table, the p-value of z = -0.68 is of 0.2483.

The p-value of the test is 0.2483 > 0.1, which means that the data does not provide sufficient evidence to conclude that the proportion of times that luggage is returned within 24 hours is less than 0.88.

7 0

3 years ago

Jose bought a magazine for $6 and five notepads. He spent a total of $26. How much did each notepad cost?

bixtya [17]

4 dollars. 26-6=20. 20/5=4

8 0

3 years ago

Find the coefficient of the x^3 in the expansion of (2x-9)^5

il63 [147K]

Use the binomial theorem:

$\displaystyle (2x-9)^5 = \sum_{k=0}^5 \binom5k (2x)^{5-k}(-9)^k = \sum_{k=0}^5 \frac{5!}{k!(5-k)!} 2^5 \left(-\frac92\right)^k x^{5-k}$