What is the surface area of the sphere shown below with a radius of 6?

WILL GIVE BRAINIEST FOR CORRECT ANSWER

lapo4ka [179]

If the center of dilation is A, this means that point A is invariant.

This means you can eliminate C and D.

Also, the scale factor is 3/4, which means that A'C' is more than half the length of AC.

This is best reflected by option B

7 0

2 years ago

The midpoint of AB is M(1, 1). If the coordinates of A are (3, -3), what are the coordinates of b

Black_prince [1.1K]

Answer:

Midpoint is ( Average of x’s, Average of y’s )

1 = (x + 7)/2

2 = x + 7, x = - 5

and - 1 = (y + (-4))/2

so - 2 = y - 4 and y = 2

Pt B = (- 5, 2)

6 0

2 years ago

Join this meeting and ill let you in (dont show face or voice and link is in replies)

patriot [66]

Answer:

yes.

Step-by-step explanation:

sike.

3 0

2 years ago

Read 2 more answers

Maria bikes 15 miles per hour and starts at 10 miles. Glen bikes 20 miles per hour and starts at 0 mile. At what time and distan

Ilya [14]

10/(20-15)=2 hours
20*2=40 miles
15*2+10=40 miles
Glen catch up with Maria after 2 hours at 40 miles

6 0

3 years ago

5) In a certain supermarket, a sample of 60 customers who used a self-service checkout lane averaged 5.2 minutes of checkout tim

Ne4ueva [31]

Answer:

$\S^2_p =\frac{(60-1)(3.1)^2 +(72 -1)(2.8)^2}{60 +72 -2}=8.643$

$S_p=2.940$

$t=\frac{(5.2 -6.1)-(0)}{2.940\sqrt{\frac{1}{60}+\frac{1}{72}}}=-1.751$

$df=60+72-2=130$

$p_v =P(t_{130}$

Assuming a significance level of $\alpha=0.05$ we have that the p value is lower than this significance level so then we can conclude that the mean for checkout time is significantly less for people who use the self-service lane

Step-by-step explanation:

Data given

Our notation on this case :

$n_1 =60$ represent the sample size for people who used a self service

$n_2 =72$ represent the sample size for people who used a cashier

$\bar X_1 =5.2$ represent the sample mean for people who used a self service

$\bar X_2 =6.1$ represent the sample mean people who used a cashier

$s_1=3.1$ represent the sample standard deviation for people who used a self service

$s_2=2.8$ represent the sample standard deviation for people who used a cashier

Assumptions

When we have two independent samples from two normal distributions with equal variances we are assuming that

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

The statistic is given by:

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

And t follows a t distribution with $n_1+n_2 -2$ degrees of freedom and the pooled variance $S^2_p$ is given by this formula:

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

System of hypothesis

Null hypothesis: $\mu_1 \geq \mu_2$

Alternative hypothesis: $\mu_1 < \mu_2$

This system is equivalent to:

Null hypothesis: $\mu_1 - \mu_2 \geq 0$

Alternative hypothesis: $\mu_1 -\mu_2 < 0$

We can find the pooled variance:

$\S^2_p =\frac{(60-1)(3.1)^2 +(72 -1)(2.8)^2}{60 +72 -2}=8.643$

And the deviation would be just the square root of the variance:

$S_p=2.940$

The statistic is given by:

$t=\frac{(5.2 -6.1)-(0)}{2.940\sqrt{\frac{1}{60}+\frac{1}{72}}}=-1.751$

The degrees of freedom are given by:

$df=60+72-2=130$

And now we can calculate the p value with:

$p_v =P(t_{130}$

Assuming a significance level of $\alpha=0.05$ we have that the p value is lower than this significance level so then we can conclude that the mean for checkout time is significantly less for people who use the self-service lane

5 0

3 years ago