If two dice are rolled one time, find the probability of getting doubles or a sum of 8.

Pls answer me this all question

Scorpion4ik [409]

Answer:

a. $5 \frac{21}{40}$

b. $\frac{1}{42}$

Step-by-step explanation:

$a. \: 12 \frac{2}{5} - 6 \frac{7}{8}$

$\frac{62}{5} - \frac{55}{8}$

$\frac{62}{8} \times \frac{8}{8} - \frac{55}{8} \times \frac{5}{5}$

$\frac{62 \times 8 - 55 \times 5}{40}$

$\frac{221}{40}$

$5 \frac{21}{40}$

$b. \: \frac{3}{14} - \frac{4}{21}$

$\frac{3}{14} \times \frac{3}{3} - \frac{4}{21} \times \frac{2}{2}$

$\frac{3 \times 3}{42} - \frac{4 \times 2}{42}$

$\frac{1}{42}$

Hope it is helpful....

5 0

3 years ago

Read 2 more answers

Monique is factoring the expression 4 x + 16 x y. Her work is shown below.

pickupchik [31]

Answer:

Step-by-step explanation:

4x + 16xy

GCF is 4x

4x(1 + 4y) <===== this is the correct one

4 0

4 years ago

Read 2 more answers

What is the approximate solution to this equation?

Nookie1986 [14]

Answer:

C. x ≈ 0.28

Step-by-step explanation:

A graphing calculator solution is attached. It shows ...

x ≈ 0.28

__

You can divide both sides of the given equation by 5, then take natural logs and divide by 5 again.

5e^(5x) = 20 . . . . . . given

e^(5x) = 4 . . . . . . . . divide by 5

5x = ln(4) . . . . . . . . take natural logs

x = ln(4)/5 . . . . . . . divide by 5

x ≈ 0.277259 ≈ 0.28

7 0

3 years ago

the perimeter of a rectangle is 128 inches. The width of the rectangle is 4y inches. Find the lent of the rectangle in terms of

ioda

Since the formula for a rectangle is lenght*2 + widht*2 then its gonna be
16 is the lenght and 4 is the width

4 0

4 years ago

a study of nickels showed that the standard deviation of the weight of nickels is 300 milligrams. A coin counter manufacture wis

cupoosta [38]

Using the z-distribution, a sample of 609 nickels has to be weighed.

<h3>What is a z-distribution confidence interval?</h3>

The confidence interval is:

$\overline{x} \pm z\frac{\sigma}{\sqrt{n}}$

The margin of error is:

$M = z\frac{\sigma}{\sqrt{n}}$

In which:

$\overline{x}$ is the sample mean.

z is the critical value.

n is the sample size.

$\sigma$ is the standard deviation for the population.

For this problem, the parameters are:

$M = 20, \sigma = 300, z = 1.645$

We solve for n to find the sample size, then:

$M = z\frac{\sigma}{\sqrt{n}}$

$20 = 1.645\frac{300}{\sqrt{n}}$

$20\sqrt{n} = 1.645 \times 300$

$\sqrt{n} = 1.645 \times 15$

$(\sqrt{n})^2 = (1.645 \times 15)^2$

n = 608.9

Rounding up, a sample of 609 nickels has to be weighed.

More can be learned about the z-distribution at brainly.com/question/25890103

#SPJ1

6 0

2 years ago