What is the 4th equivalent fraction to 1/2?

A rocket is launched from a height of 3m with an initial velocity of 15 m/s at what time will the rocket be 13m from the ground?

dem82 [27]

Not sure of my answer but, velocity=delta d/delta t. t=delta d/ velocity
—> t=(13-3)/15=>10/15=2/3s

4 0

2 years ago

A medical researcher wondered if there is a significant difference between the mean birth weight of boy and girl babies. Random

Yuliya22 [10]

Answer:

b) independent samples t-test

$t=\frac{(5.92 -5.74)-(0)}{\sqrt{\frac{1.88^2}{5}}+\frac{1.81^2}{5}}=0.154$

$p_v =2*P(t_{8}>0.154) =0.881$

So with the p value obtained and using the significance level given $\alpha=0.05$ we have $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 5% of significance the mean of the group 1 (boys) is NOT significantly different than the mean for the group 2 (Girls).

Step-by-step explanation:

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

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

We can calculate the mean and deviation for eaach group with the following formulas:

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

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

$\bar X_1 =5.92$ represent the sample mean for the group 1

$\bar X_2 =5.74$ represent the sample mean for the group 2

$s_1=1.88$ represent the sample standard deviation for group 1

$s_2=1.81$ represent the sample standard deviation for group 2

If we see the alternative hypothesis we see that we are conducting a bilateral test and with independnet samples t test.

b) independent samples t-test

The statistic is given by this formula:

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

And now we can calculate the statistic:

$t=\frac{(5.92 -5.74)-(0)}{\sqrt{\frac{1.88^2}{5}}+\frac{1.81^2}{5}}=0.154$

The degrees of freedom are given by:

$df=5+5-2=8$

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

$p_v =2*P(t_{8}>0.154) =0.881$

So with the p value obtained and using the significance level given $\alpha=0.05$ we have $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 5% of significance the mean of the group 1 (boys) is NOT significantly different than the mean for the group 2 (Girls).

7 0

4 years ago

24. If Jenna spent 30 hours on the computer this week. how much total time did she spend on

Verdich [7]

Option B) 3 hours is the time spent by Jenna on the e-mail this week.

<u>Step-by-step explanation:</u>

Jenna spent 30 hours on the computer this week.

The amount of time she spends on computer to work on various fields in given in the pie-chart.

This 30 hours represents her 100% work on computer.

From the figure shown,

She spent 10% of her total work on e-mail.

Therefore, 10% of 30 is the work done in hours by her on e-mail.

⇒ (10/100) × 30

⇒ 0.1 × 30

⇒ 3 hours

∴ Option B) 3 hours is the time spent by Jenna on the e-mail this week.

7 0

3 years ago

Which ordered pair is the solution to the system of equations?

Contact [7]

$\left \{ {{2x+y=-8} \atop {x-y=-4}} \right. \\+----\\\\3x=-12\ \ \ \ |:3\\x=-4\\\\-4-y=-4\\-4+4=y\\y=0\\\\Answer\ D(-4,0)$

5 0

3 years ago

The length of a vegetable garden is 4 feet longer than its width. If the area of the garden is 140 square feet, find its dimensi

Lena [83]

Length is 14 and the width is 10

4 0

3 years ago