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3 years ago

Solve the order of PEMDAS 25-16 divided by 2 to the power of three-5+3

1 answer:

7 0

25 - 16 / 2^3 - 5 + 3
Start off by putting 2 to the third power to get 8. Then divide 16 by 8 to get 2. 25-2=23-5=18+3=21

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Identify the solutions (roots) of the function y=x2 + 2x - 3<br><br>please help

asambeis [7]

Answer:

x=1,-3

Step-by-step explanation:

7 0

3 years ago

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

BabaBlast [244]

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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Answer:

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Step-by-step explanation:

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