Bob can run 5 miles in 1.25 hours. Stan can run 3 miles in 50 minutes. Who can run faster?

The volume of a cone is represented by the formula V = 1 T?n. If the volume of a cone is 30rt and the height is 2.5, what is the

Airida [17]

Answer:

radius is 3.38cm

Step-by-step explanation:

Note: the volume of a cone is given as

$V= 1/3 \pi r^2 h$

Step one:

Given Information

volume= 30cm^3

height= 2.5cm

Required

The radius of the cone

Step two:

substituting our data into the expression for the volume of a cone we can then solve for the radius r

$30=1/3 *3.142*r^2 * 2.5\\\\30=1/3*7.855r^2\\\\30=2.62r^2\\\\$

divide both sides by 2.62

$30/2.62= r^2\\\\11.45= r^2\\\\r=\sqrt{11.45} \\\\r=3.38cm$

7 0

3 years ago

Jennifer Lopez shbsbsbsbvevshsbbsbshhshshsbsbvsbsbsnnsjsjsjsjgdbshs

Wittaler [7]

The answer is Alex Rodriguez
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5 0

3 years ago

I need help please i will give brainly

Arisa [49]

Answer:

C

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

HELP PLEASE!!!!!!! IM TIREDD AND AM STUCK ON THIS LAST QUESTION THAT IVE BEEN WORKIGN ON FOR 20 MINUTES

avanturin [10]

Answer:

1/3

Step-by-step explanation:

7 0

3 years ago

Assume that you have a sample of n 1 equals 6, with the sample mean Upper X overbar 1 equals 50, and a sample standard deviati

tigry1 [53]

Answer:

$t=\frac{(50 -38)-(0)}{7.46\sqrt{\frac{1}{6}+\frac{1}{5}}}=2.656$

$df=6+5-2=9$

$p_v =P(t_{9}>2.656) =0.0131$

Since the p value is higher than the significance level given of 0.01 we don't have enough evidence to conclude that the true mean for group 1 is significantly higher thn the true mean for the group 2.

Step-by-step explanation:

Data given

$n_1 =6$ represent the sample size for group 1

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

$\bar X_1 =50$ represent the sample mean for the group 1

$\bar X_2 =38$ represent the sample mean for the group 2

$s_1=7$ represent the sample standard deviation for group 1

$s_2=8$ represent the sample standard deviation for group 2

System of hypothesis

The system of hypothesis on this case are:

Null hypothesis: $\mu_1 \leq \mu_2$

Alternative hypothesis: $\mu_1 > \mu_2$

We are assuming that the population variances for each group are the same

$\sigma^2_1 =\sigma^2_2 =\sigma^2$

The statistic for this case is given by:

$t=\frac{(\bar X_1 -\bar X_2)-(\mu_{1}-\mu_2)}{S_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}}$

The pooled variance is:

$S^2_p =\frac{(n_1-1)S^2_1 +(n_2 -1)S^2_2}{n_1 +n_2 -2}$

We can find the pooled variance:

$S^2_p =\frac{(6-1)(7)^2 +(5 -1)(8)^2}{6 +5 -2}=55.67$

And the pooled deviation is:

$S_p=7.46$

The statistic is given by:

$t=\frac{(50 -38)-(0)}{7.46\sqrt{\frac{1}{6}+\frac{1}{5}}}=2.656$

The degrees of freedom are given by:

$df=6+5-2=9$

The p value is given by:

$p_v =P(t_{9}>2.656) =0.0131$

Since the p value is higher than the significance level given of 0.01 we don't have enough evidence to conclude that the true mean for group 1 is significantly higher thn the true mean for the group 2.

4 0

3 years ago