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A line passes through the points (9, 30) and (18, 30). Which statement is true about the line?

Arisa [49]

(9,30)(18,30)....when u have 2 sets of points and there y value is the same, then u have a horizontal line with a slope of 0

answer : A...it has a slope of 0 because the change in the y values

7 0

4 years ago

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Match the pairs of values of f(x) and g(x) with the corresponding values of h(x) if h(x) = f(x) ÷ g(x). Tiles f(x) = x2 − 9, and

vitfil [10]

For this case we must solve each of the functions.
We have then:

f (x) = x2 - 9, and g (x) = x - 3
h (x) = (x2 - 9) / (x - 3)
h (x) = ((x-3) (x + 3)) / (x - 3)
h (x) = x + 3

f (x) = x2 - 4x + 3, and g (x) = x - 3
h (x) = (x2 - 4x + 3) / (x - 3)
h (x) = ((x-3) (x-1)) / (x - 3)
h (x) = x-1

f (x) = x2 + 4x - 5, and g (x) = x - 1
h (x) = (x2 + 4x - 5) / (x - 1)
h (x) = ((x + 5) (x-1)) / (x - 1)
h (x) = x + 5

f (x) = x2 - 16, and g (x) = x - 4
h (x) = (x2 - 16) / (x - 4)
h (x) = ((x-4) (x + 4)) / (x - 4)
h (x) = x + 4

4 0

3 years ago

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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

Read the time on the clock below.

Dmitriy789 [7]

The short line is the hand the long is the minutes that may help you.
10:25

7 0

3 years ago

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What is the answer 27x^3-1=0

chubhunter [2.5K]

X=cubic root 2/27
Glad to help

6 0

4 years ago

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