-6+-4s+-2<br><br> What is the answer to this question

The points represented by the (x, y) coordinate pairs in the

77julia77 [94]

Step-by-step explanation:

For every increase in the x-value by 1,

the y-value decreases by -2.

(From 3 to 1 and from 1 to -1)

Hence, slope of line k

= Rise/Run = (-2)/1 = -2. (A)

5 0

3 years ago

There are 84 boys in the 6th grade class. If boys make up 30% of the 6th graders, how many students are in this whole 6th grade

koban [17]

<h3>Answer: 280</h3>

Work Shown:

x = number of students in the 6th grade class

30% = 30/100 = 0.30

30% of x = 0.30x this is equal to 84 since "30% of the 6th graders are boys (and there are 84 boys in sixth grade)", so we say 0.30x = 84

Divide both sides by 0.30 to isolate x

0.30x = 84

0.30x/0.30 = 84/0.30

x = 280

So there are 280 students overall in the 6th grade class.

7 0

4 years ago

Read 2 more answers

Can someone plz help me I’m being timed!!!!

nikklg [1K]

Answer:

49 and 58

Step-by-step explanation:

pls brainliest

7 0

3 years ago

Read 2 more answers

G(x) = -(x - 3)2<br> The composite function<br> (gºg)(x)

AlexFokin [52]

Answer:

just add or * them to get the answer

Step-by-step explanation:

8 0

3 years ago

An insurance company selected a random sample of 500 clients under 18 years of age and found that 180 of them had had an acciden

Butoxors [25]

Answer:

a) The pooled proportion is p=0.3.

b) P-value = 0.000078

c) Lower bound = 0.0556

d) Upper bound = 0.1644

Step-by-step explanation:

This is a hypothesis test for the difference between proportions.

The claim is that the accident proportions differ between the two age groups .

Then, the null and alternative hypothesis are:

$H_0: \pi_1-\pi_2=0\\\\H_a:\pi_1-\pi_2\neq 0$

The significance level is 0.05.

The sample 1, of size n1=500 has a proportion of p1=0.36.

$p_1=X_1/n_1=180/500=0.36$

The sample 2, of size n2=600 has a proportion of p2=0.25.

$p_1=X_1/n_1=150/600=0.25$ .

The difference between proportions is (p1-p2)=0.11.

$p_d=p_1-p_2=0.36-0.25=0.11$

The pooled proportion, needed to calculate the standard error, is:

$p=\dfrac{X_1+X_2}{n_1+n_2}=\dfrac{180+150}{500+600}=\dfrac{330}{1100}=0.3$

The standard error for the difference between proportions can now be calculated as:

The estimated standard error of the difference between means is computed using the formula:

$s_{p1-p2}=\sqrt{\dfrac{p(1-p)}{n_1}+\dfrac{p(1-p)}{n_2}}=\sqrt{\dfrac{0.3*0.7}{500}+\dfrac{0.3*0.7}{600}}\\\\\\s_{p1-p2}=\sqrt{0.00042+0.00035}=\sqrt{0.00077}=0.0277$

Then, we can calculate the z-statistic as:

$z=\dfrac{p_d-(\pi_1-\pi_2)}{s_{p1-p2}}=\dfrac{0.11-0}{0.0277}=\dfrac{0.11}{0.0277}=3.964$

This test is a two-tailed test, so the P-value for this test is calculated as (using a z-table):

$P-value=2\cdot P(t>3.964)=0.000078$

As the P-value (0.000078) is smaller than the significance level (0.05), the effect is significant.

The null hypothesis is rejected.

There is enough evidence to support the claim that the accident proportions differ between the two age groups.

If we want to calculate the bounds of a 95% confidence interval, we start by calculating the margin of error.

For a 95% CI, the critical value for z is z=1.96.

Then, the margin of error is:

$MOE=z \cdot s_{p1-p2}=1.96\cdot 0.0277=0.0544$

Then, the lower and upper bounds of the confidence interval are:

$LL=(p_1-p_2)-z\cdot s_{p1-p2} = 0.11-0.0544=0.05561\\\\UL=(p_1-p_2)+z\cdot s_{p1-p2}= 0.11+0.0544=0.16439$

The 95% confidence interval for the population mean is (0.0556, 0.1644).

5 0

3 years ago

-6+-4s+-2 What is the answer to this question