All categories

2 years ago

PLEASE HELP ME! THE QUESTION IS IN THE SCREENSHOT BELOW!! HURRY PLEASE!!

Mathematics

1 answer:

lakkis [162]2 years ago

6 0

The answer is either A or C but I can’t be sure without seeing part A

You might be interested in

Jonah found that the equation -2|8-6|-6=-12

svetoff [14.1K]

Answer:

False equation: -2|8 - 6| - 6 = -10

Step-by-step explanation:

-2|8 - 6| - 6 = -12 True or false?

-2 | 2| - 6 = -12 Substract 8 - 6 = 2

-2 × 2 - 6 = -12 |2| = 2 absolute value of 2 is 2

- 4 - 6 = -12 -2 × 2 = -4 ( negative times positive = negative)

-10 = -12 False

7 0

3 years ago

A company wants to hire the person who

Anika [276]

Liz will get hired! To do this you will solve each person’s WPM!

Liz: 648 divided by 9 which equals 72 WPM!

Brad: 816 divided by 12 which equals only 68 WPM!

72 is greater and faster than 68 which means Liz will be hired!

Please give brainliest❤️

5 0

3 years ago

Read 2 more answers

Cloud [144]

Answer: he completed he completed 103.5 push ups in 45 seconds

7 0

3 years ago

What is the value of the expression when a = 6, b = 4, and c = 8?

Y_Kistochka [10]

Answer:

a the answer is a

Step-by-step explanation:

7 0

3 years ago

There are 200 students in eleventh grade high school class. There are 40 students in the soccer team and 50 students in the bask

bekas [8.4K]

Answer:

P(A) = 0.2

P(B) = 0.25

P(A&B) = 0.05

P(A|B) = 0.2

P(A|B) = P(A) = 0.2

Step-by-step explanation:

P(A) is the probability that the selected student plays soccer.

Then:

$P(A)=\dfrac{40}{200}=0.2$

P(B) is the probability that the selected student plays basketball.

Then:

$P(B)=\dfrac{50}{200}=0.25$

P(A and B) is the probability that the selected student plays soccer and basketball:

$P(A\&B)=\dfrac{10}{200}=0.05$

P(A|B) is the probability that the student plays soccer given that he plays basketball. In this case, as it is given that he plays basketball only 10 out of 50 plays soccer:

$P(A|B)=\dfrac{P(A\&B)}{P(B)}=\dfrac{10}{50}=0.2$

P(A | B) is equal to P(A), because the proportion of students that play soccer is equal between the total group of students and within the group that plays basketball. We could assume that the probability of a student playing soccer is independent of the event that he plays basketball.

4 0

4 years ago