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2 years ago

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The total cost for attending the rodeo carnival can be found using the function f(x) = 2.25x + 18, where x is the number of food

tickets purchased. If a family
purchases between 4 and 8 food tickets, what is the domain and range of this situation?

1 answer:

Ksju [112]2 years ago

6 0

Answer:

domain is {4,5,6,7,8}, range is the other numbers in brackets

Step-by-step explanation:

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Please help, my brain just sucks rn. The area of a rectangle is 93.6 square inches. If the length of one of its sides is 5.2 in.

snow_tiger [21]

46.4 inch

Step-by-step explanation:

Area=l×b

93.6 =5.2×b

93.6/5.2 =b

So b= 18

Again,

p=2(l+b)

p=2(18+5.2)

p= 2×23.2

p=46.4

3 0

3 years ago

Two catalysts may be used in a batch chemical process. Twelve batches were prepared using catalyst 1, resulting in an average yi

aalyn [17]

Answer:

a) $t=\frac{(91-85)-(0)}{2.490\sqrt{\frac{1}{12}}+\frac{1}{15}}=6.222$

$df=12+15-2=25$

$p_v =P(t_{25}>6.222) =8.26x10^{-7}$

So with the p value obtained and using the significance level given $\alpha=0.01$ we have $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis.

b) $(91-85) -2.79 *2.490 \sqrt{\frac{1}{12} +\frac{1}{15}} =3.309$

$(91-85) +2.79 *2.490 \sqrt{\frac{1}{12} +\frac{1}{15}} =8.691$

Step-by-step explanation:

Notation and hypothesis

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:

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

This last one is an unbiased estimator of the common variance $\sigma^2$

Part a

The system of hypothesis on this case are:

Null hypothesis: $\mu_2 \leq \mu_1$

Alternative hypothesis: $\mu_2 > \mu_1$

Or equivalently:

Null hypothesis: $\mu_2 - \mu_1 \leq 0$

Alternative hypothesis: $\mu_2 -\mu_1 > 0$

Our notation on this case :

$n_1 =12$ represent the sample size for group 1

$n_2 =15$ represent the sample size for group 2

$\bar X_1 =85$ represent the sample mean for the group 1

$\bar X_2 =91$ represent the sample mean for the group 2

$s_1=3$ represent the sample standard deviation for group 1

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

First we can begin finding the pooled variance:

$\S^2_p =\frac{(12-1)(3)^2 +(15 -1)(2)^2}{12 +15 -2}=6.2$

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

$S_p=2.490$

Calculate the statistic

And now we can calculate the statistic:

$t=\frac{(91-85)-(0)}{2.490\sqrt{\frac{1}{12}}+\frac{1}{15}}=6.222$

Now we can calculate the degrees of freedom given by:

$df=12+15-2=25$

Calculate the p value

And now we can calculate the p value using the altenative hypothesis:

$p_v =P(t_{25}>6.222) =8.26x10^{-7}$

Conclusion

So with the p value obtained and using the significance level given $\alpha=0.01$ we have $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis.

Part b

For this case the confidence interval is given by:

$(\bar X_1 -\bar X_2) \pm t_{\alpha/2} S_p \sqrt{\frac{1}{n_1} +\frac{1}{n_2}}$

For the 99% of confidence we have $\alpha=1-0.99 = 0.01$ and $\alpha/2 =0.005$ and the critical value with 25 degrees of freedom on the t distribution is $t_{\alpha/2}= 2.79$

And replacing we got:

$(91-85) -2.79 *2.490 \sqrt{\frac{1}{12} +\frac{1}{15}} =3.309$

$(91-85) +2.79 *2.490 \sqrt{\frac{1}{12} +\frac{1}{15}} =8.691$

7 0

3 years ago

Can someone please help with this!

Sever21 [200]

mathpapa should n could help

8 0

3 years ago

Read 2 more answers

What is m∠C?<br><br> Round the value to the nearest degree.

Elis [28]

Answer: $m\angle C=53\°$

Step-by-step explanation:

For this exercise you need to use the Inverse Trigonometric function arcsine, which is defined as the inverse function of the sine.

Then, to find an angle α, this is:

$\alpha =arcsin(\frac{opposite}{hypotenuse})$

In this case, you can identify that:

$\alpha =m\angle C\\\\opposite=AB=36\ cm\\\\hypotenuse=AC=45\ cm$

Then, substituting values into $\alpha =arcsin(\frac{opposite}{hypotenuse})$ and evaluating, you get that the measure of the angle "C" to the nearest degree, is:

$m\angle C=arcsin(\frac{36\ cm}{45\ cm})\\\\m\angle C=53\°$

5 0

3 years ago

What is the slope of the line that passes through (3, -4) and (-2, 5). A)-5/9 B)-9/5 C)9/5 D)5/9

alisha [4.7K]

M = (y2 - y1) / (x2 - x1)
Call (3,-4) one point
Call (-2,5) the second point.

m = (5- -4) / (-2 - 3)
m = 9/-5 = - 9 / 5

B<<<< answer

7 0

3 years ago