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nexus9112 [7]
3 years ago
8

Mr. Drumm is purchasing tickets for a school museum visit. The Museum of Natural History charges a $25 group fee plus $5 per stu

dent. The Museum of Modern Art charges a group fee of $35 plus $4 per student. To pay the same amount at each museum, how many tickets will Mr. Drumm have to order?​
Mathematics
1 answer:
Greeley [361]3 years ago
4 0

Answer:

10 tickets

Step-by-step explanation:

25 + 5x = 35 + 4x

solve for x:

subtract 25 from both sides...

5x = 35 - 25 + 4x

subtract 4x from both sides...

5x - 4x = 10

x = 10, 10 tickets.

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None of the points share a point on the x axis meaning none of them are directly vertical of each other they are all seperate from each other.

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A professor of women's studies is interested in determining if stress affects the menstrual cycle. Ten women are randomly sample
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Answer:

Null hypothesis: \mu_1 = \mu_2

Alternative hypothesis: \mu_1 \neq \mu_2

t=\frac{(20.4 -27.6)-(0)}{\sqrt{\frac{2.074^2}{5}}+\frac{2.702^2}{5}}=-4.73

df=5+5-2=8

t_{crit}=\pm 2.306

Reject H0: stress affects the menstrual cycle

Step-by-step explanation:

The statistic is given by the following formula:

t=\frac{(\bar X_1 -\bar X_2)-(\mu_{1}-\mu_2)}{\sqrt{\frac{s^2_1}{n_1}}+\frac{s^2_2}{n_2}}

Where t follows a t distribution with n_1+n_2 -2 degrees of freedom  

The system of hypothesis on this case are:

Null hypothesis: \mu_1 = \mu_2

Alternative hypothesis: \mu_1 \neq \mu_2

Or equivalently:

Null hypothesis: \mu_1 - \mu_2 = 0

Alternative hypothesis: \mu_1 -\mu_2 \neq 0

Our notation on this case :

n_1 =5 represent the sample size for group 1

n_2 =2 represent the sample size for group 2

Tha data given is:

Group 1 (High stress) : 20,23,18,19,22

Group 2 (Relatively stress free): 26,31,25,26,30

We can calculate the sample mean and the sample deviation with the following formulas:

\bar X =\frac{\sum_{i=1}^n x_i}{n}

s=\sqrt{\frac{\sum_{i=1}^n (x_i -\bar X)^2}{n-1}}

\bar X_1 =20.4 represent the sample mean for the group 1

\bar X_2 =27.6 represent the sample mean for the group 2

s_1=2.074 represent the sample standard deviation for group 1

s_2=2.702 represent the sample standard deviation for group 2

And now we can calculate the statistic:

t=\frac{(20.4 -27.6)-(0)}{\sqrt{\frac{2.074^2}{5}}+\frac{2.702^2}{5}}=-4.73

Now we can calculate the degrees of freedom given by:

df=5+5-2=8

Now we can calculate the critical value since the confidence is 95% the value for the significance would be \alpha=1-0.95=0.05 and the value for \alpha/2 =0.025 and if we find a critical value on th t distribution with 8 degrees of freedom that accumulates 0.025 of the area on each tail we got t_{crit}=\pm 2.306

Since our calculated value is lower than the critical value, we have enough evidence to reject the null hypothesis. And makes sense say that the difference between the two means are different at 5% of significance.

Reject H0: stress affects the menstrual cycle

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