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3 years ago

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A gas station sells regular Faldo line for $2.39 and premium for $2.79 per gallon. The total gas sold 550 for bot types of gasol

ine in one day for a total of 1,344.50 how many gallons of premium gasoline were sold?

1 answer:

Genrish500 [490]3 years ago

5 0

2.39x+2.79y=1344.50 where x is regular and y is premium
x+y=550

U will get x=475 y=75
75 gallon premium

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Answer:

$t=\frac{(50 -38)-(0)}{7.46\sqrt{\frac{1}{6}+\frac{1}{5}}}=2.656$

$df=6+5-2=9$

$p_v =P(t_{9}>2.656) =0.0131$

Since the p value is higher than the significance level given of 0.01 we don't have enough evidence to conclude that the true mean for group 1 is significantly higher thn the true mean for the group 2.

Step-by-step explanation:

Data given

$n_1 =6$ represent the sample size for group 1

$n_2 =5$ represent the sample size for group 2

$\bar X_1 =50$ represent the sample mean for the group 1

$\bar X_2 =38$ represent the sample mean for the group 2

$s_1=7$ represent the sample standard deviation for group 1

$s_2=8$ represent the sample standard deviation for group 2

System of hypothesis

The system of hypothesis on this case are:

Null hypothesis: $\mu_1 \leq \mu_2$

Alternative hypothesis: $\mu_1 > \mu_2$

We are assuming that the population variances for each group are the same

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

The statistic for this case is given by:

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

The pooled variance is:

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

We can find the pooled variance:

$S^2_p =\frac{(6-1)(7)^2 +(5 -1)(8)^2}{6 +5 -2}=55.67$

And the pooled deviation is:

$S_p=7.46$

The statistic is given by:

$t=\frac{(50 -38)-(0)}{7.46\sqrt{\frac{1}{6}+\frac{1}{5}}}=2.656$

The degrees of freedom are given by:

$df=6+5-2=9$

The p value is given by:

$p_v =P(t_{9}>2.656) =0.0131$

Since the p value is higher than the significance level given of 0.01 we don't have enough evidence to conclude that the true mean for group 1 is significantly higher thn the true mean for the group 2.

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3 years ago

A recipe calls for 6.67 cups of sugar, but Jake only has 1.8 cups. How much

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Answer:

We conclude that the mean lead concentration for all such medicines is less than 17 mu.

Step-by-step explanation:

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<u><em>Let </em></u> $\mu$ <u><em> = mean lead concentration for all such medicines.</em></u>

So, Null Hypothesis, $H_0$ : $\mu$ = 17 mu {means that mean lead concentration for all such medicines is equal to 17 mu}

Alternate Hypothesis, $H_A$ : $\mu$ < 17 mu {means that the mean lead concentration for all such medicines is less than 17 mu}

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P-value = P( $t_9$ < -2.16) = <u>0.031</u>

<u><em>Now, at 0.10 significance level the t table gives critical value of -1.383 for left-tailed test.</em></u><em> Since our test statistics is less than the critical value of t as -2.16 < -1.383, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region due to which </em><em><u>we reject our null hypothesis</u></em><em>.</em>

Therefore, we conclude that the mean lead concentration for all such medicines is less than 17 mu.

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3 years ago