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3 years ago

10

To make fruit salad sara uses 28 ounces of pineapple 21 ounces of apples 19 ounces of bannas and 16onces of mango how many 6-oun

ce serving of fruit salaf can sarah make?

1 answer:

grandymaker [24]3 years ago

4 0

14 servings,
28+21+19+16 = 84
and 84÷6 = 14

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1. Bailey withdrew $135 from her checking account. Write an algebraic expression that describes Bailey's action.

MArishka [77]

<u><em>Answer:</em></u>

c − 135

<u><em>Explanation: </em></u>

Let's say that the amount of money in Bailey's checking account is

c dollars.

We also know that she took away $ 135

From this variable and number, we can state the algebraic expression

c − 135

<h2><u><em>Hope this helps!</em></u></h2>

<u><em></em></u>

7 0

2 years ago

A research engineer for a tire manufacturer is investigating tire life for a new rubber compound and has built 16 tires and test

frutty [35]

Answer:

The 95% confidence interval for mean tire life is between 52,368.4 kilometers and 67,911 kilometers.

Step-by-step explanation:

We are in posession of the sample's standard deviation, so we use the students t-distribution to solve this question.

The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So

df = 16 - 1 = 15

95% confidence interval

Now, we have to find a value of T, which is found looking at the t table, with 15 degrees of freedom(y-axis) and a confidence level of [tex]1 - \frac{1 - 0.95}{2} = 0.975. So we have T = 2.68

The margin of error is:

M = T*s = 2.1315*3645.94 = 7771.3

In which s is the standard deviation of the sample.

The lower end of the interval is the sample mean subtracted by M. So it is 60139.7 - 7771.3 = 52,368.4.

The upper end of the interval is the sample mean added to M. So it is 60139.7 + 7771.3 = 67,911.

The 95% confidence interval for mean tire life is between 52,368.4 kilometers and 67,911 kilometers.

3 0

2 years ago

Read 2 more answers

Trisha had the following seven distances (in feet) in the triple jump.

Murrr4er [49]

When you find the mean you add all of them together. Then you divide your answer by how many numbers there are that you added.

So what I did was add 29+30+28+32+29+31+27= 205
then divide 205 by 7 and you get 29.285

7 0

3 years ago

Read 2 more answers

WHat is the unit rate in terms of people per square mile when there are 646,449 people in 68.3 square miles?

Oliga [24]

If there are 646,449 people in 68.3 square miles. each square mile area has 9465 people.

8 0

3 years ago

Read 2 more answers

23. A certain type of gasoline is supposed to have a mean octane rating of at least 90. Suppose measurements are taken on of 5 r

Nostrana [21]

Answer:

a) $p_v =P(Z$

b) Since $p_v$ we can reject the null hypothesis at the significance level given. So based on this we can commit type of Error I.

Because Type I error " is the rejection of a true null hypothesis".

c) $P(\bar X >90)=1-P(Z$

$1-P(Z$

d) For this case we need a z score that accumulates 0.02 of the area on the right tail and 0.98 on the left tail and this value is z=2.054

And we can use this formula:

$2.054=\frac{90-89}{\frac{0.8}{\sqrt{n}}}$

And if we solve for n we got:

$n = (\frac{2.054 *0.8}{1})^2=2.7 \approx 3$

Step-by-step explanation:

Previous concepts and data given

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".

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

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

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

$\bar X=88.96$ represent the sample mean

$s=1.011$ represent the sample standard deviation

n=5 represent the sample selected

$\alpha$ significance level

State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the mean is less than 90, the system of hypothesis would be:

Null hypothesis: $\mu \geq 90$

Alternative hypothesis: $\mu < 90$

If we analyze the size for the sample is < 30 and we don't know the population deviation so is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:

$z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}}$ (1)

z-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".

Calculate the statistic

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

$z=\frac{88.96-90}{\frac{0.8}{\sqrt{5}}}=-2.907$

Part a

P-value

First we need to calculate the degrees of freedom given by:

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

Then since is a left tailed test the p value would be:

$p_v =P(Z$

Part b

Since $p_v$ we can reject the null hypothesis at the significance level given. So based on this we can commit type of Error I.

Because Type I error " is the rejection of a true null hypothesis".

Part c

We want this probability:

$P(\bar X$

The best way to solve this problem is using the normal standard distribution and the z score given by:

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

If we apply this formula to our probability we got this:

$P(\bar X >90)=1-P(Z$

$1-P(Z$

Part d

For this case we need a z score that accumulates 0.02 of the area on the right tail and 0.98 on the left tail and this value is z=2.054

And we can use this formula:

$2.054=\frac{90-89}{\frac{0.8}{\sqrt{n}}}$

And if we solve for n we got:

$n = (\frac{2.054 *0.8}{1})^2=2.7 \approx 3$

7 0

3 years ago