Write each equation in slope intercept form Underline the slope and circle the y intercept in each equation 5x + 3y = 30

29% expressed as a fraction in simplest form is?

scoundrel [369]

The Answer: 7.25/25

Hope this helps!

5 0

3 years ago

****BRAINLIEST GOES TO THE FIRST CORRECT ANSWER****

Drupady [299]

The two equations are vertical angles, which mean they are equal.

Set each equation to equal each other and solve for x.

3x-3 = 6(x-10)

Simplify the right side:

3x-3 = 6x-60

Subtract 3x from each side:

-3 = 3x - 60

Add 60 to each side:

57 = 3x

Divide both sides by 3:

x = 57 / 3

x = 19

5 0

3 years ago

A group of 30 students is part of the audience for a TV game show. The total number of people in the audience is 140. What is th

solniwko [45]

3/8 because its out of 8 contestants n I don't think it includes the 140 audience,including the 30 students being a part if it

8 0

3 years ago

What two whole numbers does √22 fall between?<br><br> What two whole numbers does √62 fall between?

RoseWind [281]

Answer:

The 2 whole numbers that fall between √22 are 4 and 5

4*4=16, 5*5=25

The 2 whole numbers that fall between √62 are 7 and 8

7*7=49, 8*8=64

Step-by-step explanation:

8 0

4 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