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2 years ago

Plz help me fast plz plz plz thank you 16 points plz plz

Mathematics

1 answer:

nikdorinn [45]2 years ago

4 0

Answer:

The correct answer is D.

Step-by-step explanation:

The commutative property of multiplication states that it does not matter in what order things are multiplied as long as all factors are multiplied.

This example shows the numbers multiplied in different orders, but still being equal.

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What is 35 = -2x - 15 ?

gtnhenbr [62]

Answer:

x=-25

Step-by-step explanation:

35=-2x-15

add 15 to both sides, cancels the 15 from the right

50=-2x

divide both sides by -2, isolates the x

-25=x

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3 years ago

"Tim's lawn is 20 meters by 30 meters. If Tim can mow 8 square meters of grass per minute, how long will it take him to mow the

frosja888 [35]

One hour and a half

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2 years ago

There are two similar right triangles. One has side lengths of 6, 8, and 10. The other has side lengths of 24, and 18. What is t

maxonik [38]

Answer:

Step-by-step explanation:

24/8 = 3

18/6= 3

10x3 = 30

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2 years ago

Predict how much money Apple will make in the year 2025?

Goshia [24]

If we project the EPS of Apple forward 10 years to 2025, using a 42% rate of growth, we would get projected EPS of $307.34 for 2025.

So in 2025, Apple will have per share earnings of $307.34.

6 0

1 year ago

Consider a large shipment of 400 refrigerators, of which 40 have defective compressors. If X is the number among 15 randomly sel

Vlad [161]

Answer:

The required probability is 0.94

Step-by-step explanation:

Consider the provided information.

There are 400 refrigerators, of which 40 have defective compressors.

Therefore N = 400 and X = 40

The probability of defective compressors is:

$\frac{40}{400}=0.10$

It is given that If X is the number among 15 randomly selected refrigerators that have defective compressors,

That means n=15

Apply the probability density function.

$P(X=x)=^nC_xp^x(1-p)^{n-x}$

We need to find P(X ≤ 3)

$P(X\leq3) =P(X=0)+P(X=1)+P(X=2)+P(X=3)\\P(X\leq3) =\frac{15!}{15!}(0.1)^0(1-0.1)^{15}+\frac{15!}{14!}(0.1)^1(1-0.1)^{14}+\frac{15!}{13!2!}(0.1)^2(1-0.1)^{13}+\frac{15!}{12!3!}(0.1)^3(1-0.1)^{12}\\$

$P(X\leq3) =0.944444369992\approx 0.94$

Hence, the required probability is 0.94

4 0

2 years ago