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3 years ago

5x10^1

1 answer:

5 0

1. 50
2. 5
3. 0.5
4. 0.05
5. 0.005

I hope you see the pattern here . When multiplying the exponent (the small number near the 10) determines the number of zeros to the RIGHT . When dividing the number determines how many zeros to the LEFT usually resulting in a decimal. This is called scientific notation.

Hope this helps!:)

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Line m passes through the points (-2,7) and (4,-5), as shown below. Which of the following could be the equation of a line that

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Slope-intercept form: y = mx + b

(m is the slope, b is the y-intercept or the y value when x = 0 --> (0, y) or the point where the line crosses through the y-axis)

For lines to be parallel, they need to have the same slope.

To find the slope (m), use the slope formula:

$m=\frac{y_2-y_1}{x_2-x_1}$ And plug in the two points

(-2, 7) = (x₁, y₁)

(4, -5) = (x₂, y₂)

$m=\frac{y_2-y_1}{x_2-x_1}$

$m=\frac{-5-7}{4-(-2)}$ (two negative signs cancel each other out and become positive)

$m=\frac{-5-7}{4+2}$

$m=\frac{-12}{6}$ Simplify the fraction

m = -2

The slope is -2, so the parallel line's slope is also -2.

Your answer is C

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3 years ago

There are 48 female performers in a dance recital. The ratio of men to women is 5:6. How many men are in the dance recital?

Rainbow [258]

Since you said men to women, there are 5 men in the dance recital and 6 women in the dance recital.

1. Look how the problem states. Ex: Ratio of men to women.

2. Look at the ratio. 5:48.

3. We know that there are 48 female performers at the dance recital.

So, there are 5 men at the dance recital.

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3 years ago

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3 years ago

Read 2 more answers

student randomly receive 1 of 4 versions(A, B, C, D) of a math test. What is the probability that at least 3 of the 5 student te

alexdok [17]

Answer:

1.2%

Step-by-step explanation:

We are given that the students receive different versions of the math namely A, B, C and D.

So, the probability that a student receives version A = $\frac{1}{4}$ .

Thus, the probability that the student does not receive version A = $1-\frac{1}{4}$ = $\frac{3}{4}$ .

So, the possibilities that at-least 3 out of 5 students receive version A are,

1) 3 receives version A and 2 does not receive version A

2) 4 receives version A and 1 does not receive version A

3) All 5 students receive version A

Then the probability that at-least 3 out of 5 students receive version A is given by,

$\frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{3}{4}\times \frac{3}{4}$ + $\frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{3}{4}$ + $\frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}$