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vlabodo [156]
3 years ago
9

A recipe for sparkling grape juice 1 1/2 quarts of sparkling water and 3/4 of grape juice. How much sparkling water woild you ne

ed to mix with 9 quarts of grape juice. How much grape juice would you need to mix with 15/4 quarts of sparkling water?
Mathematics
2 answers:
cupoosta [38]3 years ago
4 0

According to the recipe the ratio of sparkling water to grape juice = 1 1/2 : 3/4.

Let us convert 1 1/2 into improper fraction and that is 3/2.

Now, 3/2: 3/4 or 3/2 ÷ 3/4 = 3/2 × 4/3 = 2/1 or 2:1.

Therefore, sparkling water quantity is two times of the quantity of grape juice.

Now, we need to find the number of quarts of sparkling water need to mix in 9 quarts of grape juice.


<em>Since, sparkling water quantity is two times of the quantity of grape juice.</em>

<h3>Therefore, we need 2 × 9 = 18 quarts of sparkling water to mix in 9 quarts of grape juice.</h3>

And now we need to find the number of quarts of grape juice would we need to mix with 15/4 quarts of sparkling water.

<h3>Therefore, we need 1/2 × 15/4 = 15/8 quarts of grape juice to mix with 15/4 quarts of sparkling water.</h3>

snow_lady [41]3 years ago
4 0

Answer:

a) x = 18\,qt, b) y = 1\,\frac{7}{8}\,qt

Step-by-step explanation:

a) The amount of sparkling water needed is determined by simple rule of three:

x = \frac{9\,qt}{\frac{3}{4}\,qt}\times \frac{3}{2}\,qt

x = 12\times \frac{3}{2}\,qt

x = 18\,qt

b) The amount of grape juice needed is found by simple rule of three:

y = \frac{\frac{15}{4}\,qt}{\frac{3}{2}\,qt } \times \frac{3}{4}\,qt

y = \frac{30}{12}\times \frac{3}{4}\,qt

y = \frac{90}{48}\,qt

y = \frac{15}{8}\,qt

y = 1\,\frac{7}{8}\,qt

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A simple random sample of size nequals10 is obtained from a population with muequals68 and sigmaequals15. ​(a) What must be true
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(a) The distribution of the sample mean (\bar x) is <em>N</em> (68, 4.74²).

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Step-by-step explanation:

A random sample of size <em>n</em> = 10 is selected from a population.

Let the population be made up of the random variable <em>X</em>.

The mean and standard deviation of <em>X</em> are:

\mu=68\\\sigma=15

(a)

According to the Central Limit Theorem if we have a population with mean <em>μ</em> and standard deviation <em>σ</em> and we take appropriately huge random samples (<em>n</em> ≥ 30) from the population with replacement, then the distribution of the sample mean will be approximately normally distributed.

Since the sample selected is not large, i.e. <em>n</em> = 10 < 30, for the distribution of the sample mean will be approximately normally distributed, the population from which the sample is selected must be normally distributed.

Then, the mean of the distribution of the sample mean is given by,

\mu_{\bar x}=\mu=68

And the standard deviation of the distribution of the sample mean is given by,

\sigma_{\bar x}=\frac{\sigma}{\sqrt{n}}=\frac{15}{\sqrt{10}}=4.74

Thus, the distribution of the sample mean (\bar x) is <em>N</em> (68, 4.74²).

(b)

Compute the value of P(\bar X as follows:

P(\bar X

                    =P(Z

*Use a <em>z</em>-table for the probability.

Thus, the value of P(\bar X is 0.7642.

(c)

Compute the value of P(\bar X\geq 69.1) as follows:

Apply continuity correction as follows:

P(\bar X\geq 69.1)=P(\bar X> 69.1+0.5)

                    =P(\bar X>69.6)

                    =P(\frac{\bar X-\mu_{\bar x}}{\sigma_{\bar x}}>\frac{69.6-68}{4.74})

                    =P(Z>0.34)\\=1-P(Z

Thus, the value of P(\bar X\geq 69.1) is 0.3670.

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