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4 years ago

Using the image shown, name the reflection of B over the horizontal mirror line shown.

Mathematics

1 answer:

zubka84 [21]4 years ago

8 0

The answer to your question is the letter G.

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-3(r+2)=40 whats is r in the equation , thanks :)

Tom [10]

Answer:

r = -46/3

Step-by-step explanation:

-3(r+2)=40

Distribute

-3r - 6 = 40

Add 6 to each side

-3r-6+6 = 40+6

-3r = 46

Divide by -3

r = 46/-3

r = -46/3

6 0

4 years ago

1.5 divided by 12

Marta_Voda [28]

Step-by-step explanation:

1.5 ÷ 12

1½÷12

³/²÷12

3/2 *1/12

3/24

1/8

5 0

3 years ago

Read 2 more answers

Solve the following: 2(5x-3)

Otrada [13]

Answer:

The vale of x is 3/5.

Step-by-step explanation:

First you have to expand out :

$2(5x - 3)$

$= 2(5x) - 2(3)$

$= 10x - 6$

Next you have to solve it by adding 6 to both sides :

Let expression = 0,

$10x - 6 = 0$

$10x - 6 + 6 = 0 + 6$

$10x = 6$

$x = \frac{6}{10}$

$x = \frac{3}{5}$

6 0

3 years ago

Read 2 more answers

She wishes to estimate the proportion of horses with enteroliths who are fed at least two flakes of alfalfa per day. In a sample

pogonyaev

Answer:

0.0594.

Step-by-step explanation:

The standard error of a proportion p, in a sample of size n, is given by:

$SE_{p} = \sqrt{\frac{p(1-p)}{n}}$

In a sample of 62 horses with enteroliths, she finds 42 are fed two or more flakes of alfalfa.

This means that $n = 62, p = \frac{42}{62} = 0.6774$

$SE_{p} = \sqrt{\frac{p(1-p)}{n}}$

$SE_{p} = \sqrt{\frac{0.6774*0.3226}{62}} = 0.0594$

So the correct answer is:

0.0594.

6 0

3 years ago

A certain test preparation course is designed to help students improve their scores on the LSAT exam. A mock exam is given at th

nasty-shy [4]

Answer:

(9.6, 25.7) is a 80% confidence interval for the average net change in a student's score after completing the course.

Step-by-step explanation:

We have n = 6, $\bar{x} = 17.6667$ and s = 13.3367. The confidence interval is given by

$\bar{x}\pm t_{\alpha/2}(\frac{s}{\sqrt{n}})$ where $t_{\alpha/2}$ is the $\alpha/2$ th quantile of the t distribution with n-1=5 degrees of freedom. As we want the 80% confidence interval, we have that $\alpha = 0.2$ and the confidence interval is $17.6667\pm t_{0.1}(\frac{13.3367}{\sqrt{6}})$ where $t_{0.1}$ is the 10th quantile of the t distribution with 5 df, i.e., $t_{0.1} = -1.4759$ . Then, we have $17.6667\pm (1.4759)(\frac{13.3367}{\sqrt{6}})$ and the 80% confidence interval is given by (9.6, 25.7)

6 0

3 years ago