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andrew11 [14]
4 years ago
14

Mike is preparing refreshments for a party. To make a smoothie, he will mix 3 1⁄2 quarts of strawberry puree with 2 pints of lem

onade and 1 1⁄4 gallons of water. How much smoothie will he make?
Mathematics
2 answers:
Shkiper50 [21]4 years ago
7 0
Let's convert all volumes into pint:

1 Quart  = 2 pints → 3.5 quart of strawberry = 2 x 3.5 = 7 pints Strawberry
1 Gallon = 8 pints →1.25 Ga. water = 1.25 x 8 = 10 pints Water
2 pints of limonade = 2 pints of limonade 

Add up all pints Total = 7+10+2 = 19 pints of smoothie

Or in quart 9.5

Or in Gal 2.375

shtirl [24]4 years ago
5 0

Strawberry puree in pints:

3 1⁄2 quarts = (3 × 2) + (1⁄2 × 2) = 7 pints

Water in pints:

1 1⁄4 gallons = (1 × 8) + (1⁄4 × 8) = 10 pints

Total amount of smoothie in pints:

7 + 2 + 10 = 19 pints

Total amount of smoothie in gallons and pints:

19 pints = 19 ÷ 8 = 2 remainder 3 = 2 gallons 3 pints

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3 0
3 years ago
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7 0
4 years ago
Read 2 more answers
The package of a particular brand of rubber band says that the bands can hold a weight of 7 lbs. Suppose that we suspect this mi
Nadusha1986 [10]

Answer:

The statement is True.

Step-by-step explanation:

In this case we need to determine whether the rubber bands in a package of a particular brand of rubber band can hold a weight of 7 lbs or less.

A one-sample test can be used to perform the analysis.

The hypothesis can be defined as follows:

<em>H₀</em>: The mean weight the rubber bands can hold is 7 lbs, i.e. <em>μ</em> = 7.

<em>Hₐ</em>: The mean weight the rubber bands can hold is less than 7 lbs, i.e. <em>μ</em> < 7.

The information provided is:

 n=36\\\bar x=6.6\ \text{lbs}\\\sigma=2\ \text{lbs}

As the population standard deviation is provided, we will use a <em>z</em>-test for single mean.

Compute the test statistic value as follows:

z=\frac{\bar x-\mu}{\sigma/\sqrt{n}}=\frac{6.6-7}{2/\sqrt{36}}=-1.20  

The test statistic value is -1.20.

Decision rule:

If the p-value of the test is less than the significance level then the null hypothesis will be rejected.

Compute the <em>p</em>-value for the two-tailed test as follows:

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*Use a z-table for the probability.

The p-value of the test is 0.1151.

The <em>p</em>-value of the test is very large for all the commonly used significance level. The null hypothesis will not be rejected.

Thus, it can be concluded that the mean weight the rubber bands can hold is 7 lbs.

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7 0
3 years ago
HELP I NEED HELP ASAP
earnstyle [38]

Answer:

C

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Therefore letter C is the answer.

4 0
3 years ago
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