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sertanlavr [38]
1 year ago
15

Grant plans to evaporate enough water from 22 gallons of a 16% ammonia solution to make a 24% ammonia solution. Which equation c

an he use to find n, the number of gallons of water he should remove?
Mathematics
1 answer:
Paha777 [63]1 year ago
7 0
First lets write down our starting conditions:

we have 22 gallons and it has 16% of ammonia

16% of 22 gallons is
A = 16/100*22 = 3,52 gallons

Now since only watter is evaporating that means that in second scenario after some watter evaporated there is still 3,52 gallons of ammonia. But now those 3,52 gallons of ammonia represent 24% of total mixture. So now we go other way. 
24 % is 3,52 gallons
100% is x gallons

x = 100/24*3,52 = 14.667 gallons

That means that he has to evaporate 
22 - 14.6667 = 7,333 gallons of water.
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Two samples each of size 20 are taken from independent populations assumed to be normally distributed with equal variances. The
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Answer:

t=\frac{(43.5 -40.1)-(0)}{3.678\sqrt{\frac{1}{20}+\frac{1}{20}}}=2.923

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And the p value is given by:

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Since the p value for this cae is lower than the significance level of 0.05 we have enough evidence to reject the null hypothesis and we can conclude that the true means for this case are significantly different

Step-by-step explanation:

When we have two independent samples from two normal distributions with equal variances we are assuming that  

\sigma^2_1 =\sigma^2_2 =\sigma^2

And the statistic is given by this formula:

t=\frac{(\bar X_1 -\bar X_2)-(\mu_{1}-\mu_2)}{S_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}}

Where t follows a t distribution with n_1+n_2 -2 degrees of freedom and the pooled variance S^2_p is given by this formula:

S^2_p =\frac{(n_1-1)S^2_1 +(n_2 -1)S^2_2}{n_1 +n_2 -2}

The system of hypothesis on this case are:

Null hypothesis: \mu_1 = \mu_2

Alternative hypothesis: \mu_1 \neq \mu_2

We have the following data given:

n_1 =20 represent the sample size for group 1

n_2 =20 represent the sample size for group 2

\bar X_1 =43.5 represent the sample mean for the group 1

\bar X_2 =40.1 represent the sample mean for the group 2

s_1=4.1 represent the sample standard deviation for group 1

s_2=3.2 represent the sample standard deviation for group 2

First we can begin finding the pooled variance:

\S^2_p =\frac{(20-1)(4.1)^2 +(20 -1)(3.2)^2}{20 +20 -2}=13.525

And the deviation would be just the square root of the variance:

S_p=3.678

The statistic is givne by:

t=\frac{(43.5 -40.1)-(0)}{3.678\sqrt{\frac{1}{20}+\frac{1}{20}}}=2.923

The degrees of freedom are

df=20+20-2=38

And the p value is given by:

p_v =2*P(t_{38}>2.923) =0.0058

Since the p value for this cae is lower than the significance level of 0.05 we have enough evidence to reject the null hypothesis and we can conclude that the true means for this case are significantly different

6 0
1 year ago
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