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satela [25.4K]
3 years ago
11

-6(-p + 8) = -6p + 12

Mathematics
2 answers:
kolbaska11 [484]3 years ago
6 0
ANSWER:
p=5

explanation:
distribute the -6 and simplify

a_sh-v [17]3 years ago
3 0

Answer:

p = 5

Step-by-step explanation:

-6(-p+8)=-6p+12\\6p-48=-6p+12\\12p-48=12\\12p=60\\p=5

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What is 7/12 divided by 36?
Georgia [21]
7/12 = 0.84
0.84/36 = 0.0233·
3 0
3 years ago
The diameter of a circle is 6 in. Find its area to the nearest tenth.
Marrrta [24]

Answer and Step-by-step explanation:

The area formula for a Circle is:

A = \pi r^2

Diameter is equal to the radius times 2.

<u>So, divide the diameter to get the radius.</u>

6 divided by 2 equals 3.

<u>Now, plug it into the formula.</u>

A = \pi (3)^{2}

A = 9\pi

Plug this into your calculator by multiplying 9 by pi.

<u>We get the answer to be 28.3 when rounded to the nearest tenth.</u>

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8 0
2 years ago
An alarming number of U.S. adults are either overweight or obese. The distinction between overweight and obese is made on the ba
madreJ [45]

Answer:

(A) The probability that a randomly selected adult is either overweight or obese is 0.688.

(B) The probability that a randomly selected adult is neither overweight nor obese is 0.312.

(C) The events "overweight" and "obese" exhaustive.

(D) The events "overweight" and "obese" mutually exclusive.

Step-by-step explanation:

Denote the events as follows:

<em>X</em> = a person is overweight

<em>Y</em> = a person is obese.

The information provided is:

A person is overweight if they have BMI 25 or more but below 30.

A person is obese if they have BMI 30 or more.

P (X) = 0.331

P (Y) = 0.357

(A)

The events of a person being overweight or obese cannot occur together.

Since if a person is overweight they have (25 ≤ BMI < 30) and if they are obese they have BMI ≥ 30.

So, P (X ∩ Y) = 0.

Compute the probability that a randomly selected adult is either overweight or obese as follows:

P(X\cup Y)=P(X)+P(Y)-P(X\cap Y)\\=0.331+0.357-0\\=0.688

Thus, the probability that a randomly selected adult is either overweight or obese is 0.688.

(B)

Commute the probability that a randomly selected adult is neither overweight nor obese as follows:

P(X^{c}\cup Y^{c})=1-P(X\cup Y)\\=1-0.688\\=0.312

Thus, the probability that a randomly selected adult is neither overweight nor obese is 0.312.

(C)

If two events cannot occur together, but they form a sample space when combined are known as exhaustive events.

For example, flip of coin. On a flip of a coin, the flip turns as either Heads or Tails but never both. But together the event of getting a Heads and Tails form a sample space of a single flip of a coin.

In this case also, together the event of a person being overweight or obese forms a sample space of people who are heavier in general.

Thus, the events "overweight" and "obese" exhaustive.

(D)

Mutually exclusive events are those events that cannot occur at the same time.

The events of a person being overweight and obese are mutually exclusive.

5 0
2 years ago
Solve for Z! If you can explain
Bogdan [553]
The answer is Z=-3 :)
7 0
2 years ago
At a strawberry stand Luisa paid $4.50 for 2 pounds of strawberries. How much would you expect to pay for five pounds or strawbe
Daniel [21]

<u>Half</u> of 4.50 is 2.25, so for <u>one</u> pound it would be <u>$2.25</u>.

To calculate five pounds, you need to add 2.25+2.25+2.25+2.25+2.25 = <u>11.25</u>.

For <u>five</u> pounds of strawberry's, it would cost <u>$11.25</u>.

Hope this helped !

4 0
3 years ago
Read 2 more answers
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