Suppose that a normal distribution has a mean of 20 and a standard deviation of 4. What is the z-score of 13?

Given: circle k(O), m∠A=(x+45)°, m KE =(x+20)°, m EW =(3x)° Find: m KW . HELP ASAP WILL GIVE BRAINLIEST!

egoroff_w [7]

Answer:

160°

Step-by-step explanation:

We can use the central angle theorem for this. It says that an angle inscribed in a circle (on the circumference) measured half of the arc it intercepts (by 2 points on the circumference of the circle).

that would mean that Angle KAW is half of measure of Arc KW. Thus we can write:

m∠A = 0.5 (arc KE + arc EW)

x+45 =0.5 (x+20+3x)

x+45 = 0.5(4x+20)

x+45 = 2x + 10

45 - 10 = 2x - x

Thus, x = 35

Since Arc KW = Arc KE + Arc EW and x = 35, we can say:

Arc KW = x + 20 + 3x = 35 + 20 + 3(35) = 160°

6 0

3 years ago

Earth is 92,955,820 miles from the sun, Mars is 142,633,260 miles from the sun. What is the distance the martians would have to

Maru [420]

For martians to destroy human society, they would need to travel 49,667,440 miles.

4 0

3 years ago

How do you solve this algebreically? 3/8=x/20

Llana [10]

Answer:

7.5

Step-by-step explanation:

3/8 = x/20

3(20) = 8x

8x = 60

x = 60/8

x = 7.5

6 0

3 years ago

PLEASE HELP!!! 25 POINTS!!!

Phoenix [80]

6 4
5

2 3
6

7 5
3

2 4
8

3 5
7

3 0

2 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