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A runner compared how long it took him to run 200 meters to the record time held by a friend. In his first heat, his time was 2.

Rasek [7]

Answer: it took him 0.075 seconds lesser

Step-by-step explanation:

In his first heat, his time was 2.45 seconds greater than his friend's record time. In his second heat, his time was 2 3/8 seconds greater than his friend's record time. Converting 2 3/8 seconds to decimal, it becomes 2.375 seconds

Therefore, he was fastest in the second heat.

The difference between the time he used to run in the second heat than in the first heat is

2.45 - 2.375 = 0.075 seconds

4 0

3 years ago

Need Help ASAP <br> Please help with this math question i dont understand it very well.

mixas84 [53]

Answer:

-30h +40 - 60 I think.........

6 0

3 years ago

Jimmy applied the distributive property to write the equation below.<br><br> 24 + 6 = 6 (4 + 2)

joja [24]

Answer: if you simplify this it becomes 30=28 but this is a false statement so you would say 30≠28

Step-by-step explanation:

≠ means not equal

7 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

3 years ago

Find the equation of a line parallel to y - 5x = 10 that passes through the point (3, 10). (answer in slope-intercept form)

Maru [420]

I think it would be the second one

8 0

3 years ago

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