Ask question

Ask question

All categories

2 years ago

8

15. A student received a 75% on an exam. If there were 20 questions on the exam, which ratio would show the number of questions

answered correctly to the total number of questions?
O A. 15 20
O B. 20:15
O C. 20.5
O D.5.20

1 answer:

Juliette [100K]2 years ago

3 0

Answer:

75% = 3/4

3/4 of 20 = 15

Ratio - 15:20 (A)

You might be interested in

A line has a slope of 3 and a y-intercept of 5. What is its equation in slope-intercept form? Write your answer using integers,

s344n2d4d5 [400]

Answer:

$y=3x+5$

Step-by-step explanation:

The problem is asking for slope-intercept form, luckily, they gave us both of those things.

Slope-intercept form: $y=mx+b$ , where $m=$ slope and $b=$ y-intercept.

So,

$y=3x+5$

Hope this helps!

7 0

3 years ago

What is 2+7+6-8/3*5+8-4

finlep [7]

Answer: 2+7+6-8/3*5+8-4=17/3=5 2/3 =5.666667

Step-by-step explanation:

7 0

3 years ago

PLZ HURRY Last question i need a good grade

NISA [10]

Answer:

C. The lines are perpendicular

Step-by-step explanation:

GO GET GOOD GrADE

4 0

2 years ago

Write a paragraph comparing the two classes' semester grades. Be sure to compare the extremes, the quartiles, the medians, and t

Gnom [1K]

Answer:

Second class have higher marks and greater spread.

Step-by-step explanation:

First box plot represents class first. From the first box plot, we get

$\text{Minimum value }= 53,Q_1=62,Median=80,Q_3=86,\text{Maximum value }= 89$

$Range=Maximum-Minimum=89-53=36$

$IQR=Q_3-Q-1=86-62=24$

Second box plot represents class second. From the second box plot, we get

$\text{Minimum value }= 56,Q_1=62,Median=74,Q_3=89,\text{Maximum value }= 96$

$Range=Maximum-Minimum=96-56=40$

$IQR=Q_3-Q-1=89-62=27$

First class has greater minimum value, first quartile of both classes are same, second class has greater median, first class has greater third quartile and first class has greater maximum value. It means second class have higher marks but class first have less variation.

Second class has greater range and greater inter quartile range. It means data of second class has greater spread.

Therefore, second class have higher marks and greater spread.

3 0

3 years ago

Evaluate cube root of 5 multiplied by square root of 5 over cube root of 5 to the power of 5.

Andre45 [30]

Answer:

Option d) 5 to the power of negative 5 over 6 is correct.

$\dfrac{\sqrt[3]{\bf 5} \times \sqrt{\bf 5}}{\sqrt[3]{\bf 5^{\bf 5}}}= 5^{\frac{\bf -5}{\bf 6}}$

Above equation can be written as 5 to the power of negative 5 over 6.

ie, $5^\frac{\bf -5}{\bf 6}$

Step-by-step explanation:

Given that cube root of 5 multiplied by square root of 5 over cube root of 5 to the power of 5.

It can be written as below

$\dfrac{\sqrt[3]{5} \times \sqrt{5}}{\sqrt[3]{5^5}}$

$\dfrac{\sqrt[3]{5} \times \sqrt{5}}{\sqrt[3]{5^5}}= \dfrac{5^{\frac{1}{3}} \times 5^{\frac{1}{2}}}{5^{\frac{5}{3}}}$

$\dfrac{\sqrt[3]{5} \times \sqrt{5}}{\sqrt[3]{5^5}}= \dfrac{5^{\frac{1}{3}+\frac{1}{2}}}{5^{\frac{5}{3}}}$

$\dfrac{\sqrt[3]{5} \times \sqrt{5}}{\sqrt[3]{5^5}}= \dfrac{5^{\frac{2+3}{6}}}{5^{\frac{5}{3}}}$

$\dfrac{\sqrt[3]{5} \times \sqrt{5}}{\sqrt[3]{5^5}}= 5^{\frac{5}{6}} \times 5^{\frac{-5}{3}}$

$\dfrac{\sqrt[3]{5} \times \sqrt{5}}{\sqrt[3]{5^5}}= 5^{\frac{5-10}{6}}$

$\dfrac{\sqrt[3]{5} \times \sqrt{5}}{5^5}= 5^{\frac{-5}{6}}$

Above equation can be written as 5 to the power of negative 5 over 6.

7 0

3 years ago