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vivado [14]
3 years ago
10

A 20-gallon salt-water solution contains 15% pure salt. How much pure water should be added to it to produce a 10% solution?

Mathematics
1 answer:
Elanso [62]3 years ago
3 0

15% of 20 gal = 0.15 * 20 gal = 3 gal, so the solution contains 3 gal of salt.

If we add <em>x</em> gal of water to the solution, we end up with (20 + <em>x</em>) gal of solution. We want the new mixture to have a concentration of 10%, or

10% of (20 + <em>x</em>) gal = 0.1 * (20 + <em>x</em>) gal = 2 + 0.1<em>x</em> gal

of salt.

The amount of salt in the tank hasn't changed. Solve for <em>x </em>:

2 + 0.1<em>x</em> = 3

0.1<em>x</em> = 1

<em>x</em> = 10 gal

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There is not enough evidence to support the claim that the liquid diet yields a higher mean weight loss than the powder diet (P-value = 0.15).

Step-by-step explanation:

This is a hypothesis test for the difference between populations means.

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Then, the null and alternative hypothesis are:

H_0: \mu_1-\mu_2=0\\\\H_a:\mu_1-\mu_2< 0

The significance level is 0.05.

The sample 1 (powder diet group), of size n1=49 has a mean of 42 and a standard deviation of 12.

The sample 2 (liquid diet group), of size n2=36 has a mean of 45 and a standard deviation of 14.

The difference between sample means is Md=-3.

M_d=M_1-M_2=42-45=-3

The estimated standard error of the difference between means is computed using the formula:

s_{M_d}=\sqrt{\dfrac{\sigma_1^2}{n_1}+\dfrac{\sigma_2^2}{n_2}}=\sqrt{\dfrac{12^2}{49}+\dfrac{14^2}{36}}\\\\\\s_{M_d}=\sqrt{2.939+5.444}=\sqrt{8.383}=2.895

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t=\dfrac{M_d-(\mu_1-\mu_2)}{s_{M_d}}=\dfrac{-3-0}{2.895}=\dfrac{-3}{2.895}=-1.04

The degrees of freedom for this test are:

df=n_1+n_2-1=49+36-2=83

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\text{P-value}=P(t

As the P-value (0.15) is greater than the significance level (0.05), the effect is not significant.

The null hypothesis failed to be rejected.

There is not enough evidence to support the claim that the liquid diet yields a higher mean weight loss than the powder diet.

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