Ask question

Ask question

All categories

MA_775_DIABLO [31]

2 years ago

11

Mr. Burns organized the grades for 11 students in the boxplot below. None of the students earned the same grade. What percent of

grades are between 79 and 84?
A. 100%

B. 25%

C. 50%

D. 75%

1 answer:

leva [86]2 years ago

6 0

Answer:

C

Step-by-step explanation:

You might be interested in

which of the following is equivalent to 3 sqrt 32x^3y^6 / 3 sqrt 2x^9y^2 where x is greater than or equal to 0 and y is greater

Nutka1998 [239]

Answer:

$\frac{\sqrt[3]{16y^4}}{x^2}$

Step-by-step explanation:

The options are missing; However, I'll simplify the given expression.

Given

$\frac{\sqrt[3]{32x^3y^6}}{\sqrt[3]{2x^9y^2} }$

Required

Write Equivalent Expression

To solve this expression, we'll make use of laws of indices throughout.

From laws of indices $\sqrt[n]{a} = a^{\frac{1}{n}}$

So,

$\frac{\sqrt[3]{32x^3y^6}}{\sqrt[3]{2x^9y^2} }$ gives

$\frac{(32x^3y^6)^{\frac{1}{3}}}{(2x^9y^2)^\frac{1}{3}}$

Also from laws of indices

$(ab)^n = a^nb^n$

So, the above expression can be further simplified to

$\frac{(32^\frac{1}{3}x^{3*\frac{1}{3}}y^{6*\frac{1}{3}})}{(2^\frac{1}{3}x^{9*\frac{1}{3}}y^{2*\frac{1}{3}})}$

Multiply the exponents gives

$\frac{(32^\frac{1}{3}x*y^{2})}{(2^\frac{1}{3}x^{3}*y^{\frac{2}{3}})}$

Substitute $2^5$ for 32

$\frac{(2^{5*\frac{1}{3}}x*y^{2})}{(2^\frac{1}{3}x^{3}*y^{\frac{2}{3}})}$

$\frac{(2^{\frac{5}{3}}x*y^{2})}{(2^\frac{1}{3}x^{3}*y^{\frac{2}{3}})}$

From laws of indices

$\frac{a^m}{a^n} = a^{m-n}$

This law can be applied to the expression above;

$\frac{(2^{\frac{5}{3}}x*y^{2})}{(2^\frac{1}{3}x^{3}*y^{\frac{2}{3}})}$ becomes

$2^{\frac{5}{3}-\frac{1}{3}}x^{1-3}*y^{2-\frac{2}{3}}$

Solve exponents

$2^{\frac{5-1}{3}}*x^{-2}*y^{\frac{6-2}{3}}$

$2^{\frac{4}{3}}*x^{-2}*y^{\frac{4}{3}}$

From laws of indices,

$a^{-n} = \frac{1}{a^n}$ ; So,

$2^{\frac{4}{3}}*x^{-2}*y^{\frac{4}{3}}$ gives

$\frac{2^{\frac{4}{3}}*y^{\frac{4}{3}}}{x^2}$

The expression at the numerator can be combined to give

$\frac{(2y)^{\frac{4}{3}}}{x^2}$

Lastly, From laws of indices,

$a^{\frac{m}{n} = \sqrt[n]{a^m}$ ; So,

$\frac{(2y)^{\frac{4}{3}}}{x^2}$ becomes

$\frac{\sqrt[3]{(2y)}^{4}}{x^2}$

$\frac{\sqrt[3]{16y^4}}{x^2}$

Hence,

$\frac{\sqrt[3]{32x^3y^6}}{\sqrt[3]{2x^9y^2} }$ is equivalent to $\frac{\sqrt[3]{16y^4}}{x^2}$

8 0

3 years ago

What is the unit rate for 98.4 miles on 4.8 gallons of gas?<br><br> 2.05<br> 20.5<br> 25<br> 205

igor_vitrenko [27]

20.5 is the answer because when you divide 98.4 by 4.8 you get 20.5 as the answer

3 0

2 years ago

8.) It takes Fred 9 minutes to type and spell check 13 pages of a manuscript. Find how long it takes him to type and spell check

inysia [295]

It will take Fred 42 minutes to type and spell check 60 pages.

Let the first scenario be A.

Let the second scenario be B.
Let the unknown time be X.

Given the following data;

Time taken A = 9 minutes

Number of pages A = 13 pages

Number of pages B = 60 pages

Time taken B = X minutes

To find how long it takes him to type and spell check 60 pages;

We would use direct proportion to express the work described and solve for the unknown variable (time).

$9 \;minutes : 13 \;pages\\\\X \;minutes : 60 \;pages$

Cross-multiplying, we have;

$9$ × $60$ $= 13$ × $X$

$540 = 13X\\\\X = \frac{540}{13}$

<em>X = 41.54 ≈ 42 minutes</em>

Therefore, it will take Fred 42 minutes to type and spell check 60 pages.

Find more information: brainly.com/question/89674

7 0

3 years ago

A student takes an exam containing 1818 true or false questions. If the student guesses, what is the probability that he will ge

creativ13 [48]

Probability

p=66×100/1818=3.6303630363%

After rounding to 4 decimals:

p=3.6304%

8 0

2 years ago

Read 2 more answers

Find the point on the parabola y^2 = 8x at which the ordinate is double the abscissa.

aliya0001 [1]

Answer:

Step-by-step explanation:

Let's make a complex of functions

y^2 = 8x

y = 2x

Now let's solve it

(2x)^2 = 8x

4x^2 - 8x = 0

4x(x - 2) = 0

x = 0, 2

y = 0, 4

So the answer is (2, 4)

Hope you like it, study well

3 0

2 years ago