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3 years ago

5

There are 25 computers in a math lab. There are 6 activity books for every 2 computers in the math lab. How many activity bookle

ts are in the math lab?

2 answers:

kondaur [170]3 years ago

6 0

Answer:

b

Step-by-step explanation:

Kazeer [188]3 years ago

4 0

Answer:

it would be A because you would have 2 books for every 6 computers and A is the only one that has that.

Step-by-step explanation:

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Please help it’s due today. I just need to finish fast. Sorry about my handwriting.

puteri [66]

Answer:

all work is shown and pictured

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3 years ago

Only I need help with 1,2 and three please

Mazyrski [523]

Is that the whole thing?

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3 years ago

Solve −18=6+6x .<br><br><br><br> x = __

Igoryamba

Answer:

x = -4

Step-by-step explanation:

-18 = 6+6x

-18 -6 = 6x

-24 = 6x

-24/6 = x

x = -4

Checking Part:

-18 = 6 + 6x

-18 = 6 + 6(-4)

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3 years ago

Read 2 more answers

Devin volunteers at Big Pup Rescue. He is mixing up shampoo solution for the dog washing station. He needs 2 cups of shampoo con

Naddika [18.5K]

Answer:

8 cups of shampoo concentrate

Step-by-step explanation:

let x=how much shampoo concentrate Devin should add

$\frac{2}{9} =\frac{x}{36}$

cross multiply

2*36=9x

72=9x

72/9=x

x=8

Devin should add 8 cups of shampoo concentrate to the water to make the shampoo solution.

4 0

3 years ago

A group of freshmen at a local university consider joining the equestrian team. Thirty‑five percent of students choose Western r

lesantik [10]

Answer:

a) Option D) 0.75

b) Option D) 0.3

Step-by-step explanation:

We are given the following in the question:

Percentage of students who choose Western riding = 35%

$P(w) = 0.35$

Percentage of students who choose dressage= 45%

$P(d) = 0.45$

Percentage of students who choose jumping = 50%

$P(j) = 0.50$

Percentage of students who choose both dressage and jumping = 20%

$P(d \cap j) = 0.20$

Percentage of students who choose Western and dressage = 10%

$P(w \cap d) = 0.10$

Percentage of students who choose Western and jumping = 0%

$P(w \cap j) = 0$

Thus, we can say

$P(w \cap j \cap d) = 0$

Formula:

$P(A \cup B) = P(A) + P(B) - P(A \cap B)$

a) P(student chooses dressage or jumping)

$P(d \cup j) = P(d) + P(j) -p(d \cap j)\\= 0.45 + 0.50-0.20 = 0.75$

b) P(student chooses neither dressage nor Western riding)

$= 1 - P(d \cup w)\\= 1 - (P(d) + P(w) - P(d \cap w))\\= 1 - (0.45 + 0.35 - 0.10)= 0.3$

5 0

3 years ago