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Marizza181 [45]

3 years ago

12

The total cost for 25 students to go to the zoo is $56. The total cost for 25 students to go to the science museum is $48.25. Ho

w much money would the school save by going to the museum?

1 answer:

Daniel [21]3 years ago

6 0

Answer:

$7.75

Step-by-step explanation:

You would subtract 48.25 from 56 (56-48.25) to get the answer.

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An auto company claims that the fuel efficiency of its sedan has been substantially improved. A consumer advocate organization w

zzz [600]

Answer:

$t=\frac{\bar d -0}{\frac{s_d}{\sqrt{n}}}=\frac{0.983 -0}{\frac{1.685}{\sqrt{12}}}=2.021$

$df=n-1=12-1=11$

$p_v =P(t_{(11)}>2.021) =0.0342$

If we compare the the p value with the significance level provided $\alpha=0.1$ , we see that $p_v < \alpha$ , so then we can reject the null hypothesis. and there is a significant increase in the miles per gallon from 2017 to 2019 at 10% of significance.

Step-by-step explanation:

Previous concepts

A paired t-test is used to compare two population means where you have two samples in which observations in one sample can be paired with observations in the other sample. For example if we have Before-and-after observations (This problem) we can use it.

Let put some notation :

x=test value 2017 , y = test value 2019

x: 28.7 32.1 29.6 30.5 31.9 30.9 32.3 33.1 29.6 30.8 31.1 31.6

y: 31.1 32.4 31.3 33.5 31.7 32.0 31.8 29.9 31.0 32.8 32.7 33.8

Solution to the problem

The system of hypothesis for this case are:

Null hypothesis: $\mu_y- \mu_x \leq 0$

Alternative hypothesis: $\mu_y -\mu_x >0$

Because if we have an improvement we expect that the values for 2019 would be higher compared with the values for 2017

The first step is calculate the difference $d_i=y_i-x_i$ and we obtain this:

d: 2.4, 0.3, 1.7,3,-0.2, 1.1, -0.5, -3.2, 1.4, 2, 1.6, 2.2

The second step is calculate the mean difference

$\bar d= \frac{\sum_{i=1}^n d_i}{n}= \frac{11.8}{12}=0.983$

The third step would be calculate the standard deviation for the differences, and we got:

$s_d =\frac{\sum_{i=1}^n (d_i -\bar d)^2}{n-1} =1.685$

We assume that the true difference follows a normal distribution. The 4th step is calculate the statistic given by :

$t=\frac{\bar d -0}{\frac{s_d}{\sqrt{n}}}=\frac{0.983 -0}{\frac{1.685}{\sqrt{12}}}=2.021$

The next step is calculate the degrees of freedom given by:

$df=n-1=12-1=11$

Now we can calculate the p value, since we have a right tailed test the p value is given by:

$p_v =P(t_{(11)}>2.021) =0.0342$

If we compare the the p value with the significance level provided $\alpha=0.1$ , we see that $p_v < \alpha$ , so then we can reject the null hypothesis. and there is a significant increase in the miles per gallon from 2017 to 2019 at 10% of significance.

4 0

3 years ago

A survey of 76 commercial airline flights of under 2 hours resulted in a sample average late time for a flight of 2.55 minutes.

nika2105 [10]

Answer:

The best point of estimate for the true mean is:

$\hat \mu = \bar X = 2.55$

$2.55-1.96\frac{12}{\sqrt{76}}=-0.148$

$2.55+1.96\frac{12}{\sqrt{76}}=5.248$

Since the time can't be negative a good approximation for the confidence interval would be (0,5.248) minutes. The interval are tellling to us that at 95% of confidence the average late time is lower than 5.248 minutes.

Step-by-step explanation:

Information given

$\bar X=2.55$ represent the sample mean for the late time for a flight

$\mu$ population mean

$\sigma=12$ represent the population deviation

n=76 represent the sample size

Confidence interval

The best point of estimate for the true mean is:

$\hat \mu = \bar X = 2.55$

The confidence interval for the true mean is given by:

$\bar X \pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$ (1)

The Confidence level given is 0.95 or 95%, th significance would be $\alpha=0.05$ and $\alpha/2 =0.025$ . If we look in the normal distribution a quantile that accumulates 0.025 of the area on each tail we got $z_{\alpha/2}=1.96$

Replacing we got:

$2.55-1.96\frac{12}{\sqrt{76}}=-0.148$

$2.55+1.96\frac{12}{\sqrt{76}}=5.248$

Since the time can't be negative a good approximation for the confidence interval would be (0,5.248) minutes. The interval are tellling to us that at 95% of confidence the average late time is lower than 5.248 minutes.

7 0

3 years ago

8x55= hmmm what is this summ

Evgen [1.6K]

Answer:

440

Step-by-step explanation:

55x8. 8x5=40 carry the 4. 8x5=40+4=44 so it is 440

5 0

3 years ago

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Plzzz hellppp URGENTLY !!!!!!5,000 rounded to the nearest hundred.

irina [24]

5000 rounded to the nearest hundredth is 5000

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The boxerville fighting hens's record has improved every year. The 2000 team won 3 more games than 1999 team and the 2001 team w

padilas [110]

1999 Team won - 15
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3 years ago

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