Solve for x 2m-nx=x+4

Step-by-step explanation: Lets first figure out how much money a box of candy bars is worth. We know that each candy bar is worth $5 and that there are 7 candy bars per box. We can multiply 7*5 to figure out that each box is worth $35.

Now lets figure out how many boxes can be sold each day. Each student can sell two boxes a day. 5 students selling 2 boxes each is 5*2= 10 boxes total sold a day.

Now finally we use what we found to figure out the amount of money made per day. We know that there are 10 boxes being sold and that each box makes them $35. We just multiply 10 boxes by $35 to get $350 dollars each day.

Since they sell for 8 days we just multiply 8*$350 to get $2800.

8 0

3 years ago

What is 3,4,5 plz and thank you

Tcecarenko [31]

For 3 it’s b. For 2 it’s 8. For 4 it’s the whole number

8 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:

X = a person is overweight

Y = 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