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Slav-nsk [51]
2 years ago
6

Solve the equation

r 8" alt="8 x = 12 power 8" align="absmiddle" class="latex-formula">
​
Mathematics
1 answer:
almond37 [142]2 years ago
6 0

Answer:

Step-by-step explanation:

8x = 12^8

x = 12*8/8

= 53747712.

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What’s the value of this expression?<br><br> 6x4-(5+3) dividend by 2
solong [7]

Answer:

20

Step-by-step explanation:

If you use PEMDAS(Parentheses, Exponents, Multiplication, Division, Addition, Subtraction) you would take care of what's in the parentheses, so you would at 5 and 3 to get 8. Then you would multiply 6 and 4 to get 24, so far your equation looks like this 24 - 8 ÷ 2. You then divide 8 by 2 to get 4 then subtract 24 from 4 to get 20.

8 0
3 years ago
Which of the following is a property of set?
NARA [144]

Answer:

C might be the answer, I guesss

8 0
3 years ago
At Feet-R-Us colored socks are $5 and white socks are $3. Jimmy buys $32 worth of socks. An equation representing this scenario
SOVA2 [1]

c=4 w=4

5c=20

3w=12

5c+3w=32

5 0
3 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
3 years ago
I need help with this please
Vitek1552 [10]
The answer to 8a. is 17, the answer to 8b. is 3.9, the answer to 8c. is 1 / 2, the answer to 8d. is 7 7 / 8, the answer to 9a. is commutative property, the answer to 9b. is multiplicative identity property, the answer to 9c. is associative property, and the answer to 9d. is additive inverse property.

Regarding the whole "explain using mental math" thing, I pretty much just used the fact that 8a. and 8b. both had the same number of decimal places. For 8c., I just made each number have a denominator of 10. And last, for 8d., all of the denominators were already the same, which made it pretty easy. I apologize for this section of my answer being so informal lol, I haven't had to do these kinds of problems in like 5 years <span />
7 0
3 years ago
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