The probability of a student getting an A in mathematics

Figure TQRS~ BCDE.Name a pair of Corresponding sides

Elis [28]

Answer:

the pair of corresponding sides are:QT=CB,RS=DE,QR=CD,TS=BE..

These are the name of corresponding sides.

hope this will help you more..good luck

4 0

3 years ago

Which equation could represent the relationship shown in the scatter plot?

horsena [70]

Answer:

A

Step-by-step explanation:

Use POE, none of the answer makes any sense

8 0

3 years ago

Read 2 more answers

The equation 3s equals 21 can be used to find the number of students s and each van on a field trip how many students are in eac

zubka84 [21]

3s=21
s=7
7 students are in each van.

6 0

4 years ago

48 ÷ [(13 – 6) — 6°] + 4 - 2 =

kvasek [131]

Answer:

brainlest plzzzzzz 50

Step-by-step explanation:

pemdas

so u you do brackets first

13-6=7

7-6=1

next you do multiplication and division

48/1=48

next is addition and subtraction

48+4=52

52-2=50

so you awnswer is 50

6 0

3 years ago

A sample of 200 observations from the first population indicated that X1 is 170. A sam- ple of 150 observations from the second

nikitadnepr [17]

Answer:

a. If the P-value is smaller than the significance level, the null hypothesis is rejected.

b. Pooled proportion = 0.8

c. z = 2.7

d. As the P-value (0.0072) is smaller than the significance level (0.05), the null hypothesis is rejected.

There is enough evidence to support the claim that the proportions differ significantly.

Step-by-step explanation:

This is a hypothesis test for the difference between proportions.

We will use the P-value approach, so the decision rule is that if the P-value is lower than the significance level, the null hypothesis is rejected.

The claim is that the proportions differ significantly.

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=200 has a proportion of p1=0.85.

$p_1=X_1/n_1=170/200=0.85$

The sample 2, of size n2=150 has a proportion of p2=0.7333.

$p_2=X_2/n_2=110/150=0.7333$

The difference between proportions is (p1-p2)=0.1167.

$p_d=p_1-p_2=0.85-0.7333=0.1167$

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

$p=\dfrac{X_1+X_2}{n_1+n_2}=\dfrac{170+110}{200+150}=\dfrac{280}{350}=0.8$

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

$s_{p1-p2}=\sqrt{\dfrac{p(1-p)}{n_1}+\dfrac{p(1-p)}{n_2}}=\sqrt{\dfrac{0.8*0.2}{200}+\dfrac{0.8*0.2}{150}}\\\\\\s_{p1-p2}=\sqrt{0.0008+0.00107}=\sqrt{0.00187}=0.0432$

Then, we can calculate the z-statistic as:

$z=\dfrac{p_d-(\pi_1-\pi_2)}{s_{p1-p2}}=\dfrac{0.1167-0}{0.0432}=\dfrac{0.1167}{0.0432}=2.7$

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

$P-value=2\cdot P(z>2.7)=0.0072$

As the P-value (0.0072) 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 proportions differ significantly.

6 0

3 years ago