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MAVERICK [17]
3 years ago
6

Please help it’s due in like 10 minutes, i’ll give you brain

Mathematics
2 answers:
neonofarm [45]3 years ago
8 0

Answer:

I think it's B, I might be wrong though

Marta_Voda [28]3 years ago
5 0

Answer:

A

Step-by-step explanation:

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3 integers less than 25 range of 10 mean of 13​
I am Lyosha [343]

Answer:

8, 13, 18

Step-by-step explanation:

If we want the mean to be 13, and there are 3 integers, that means the sum of all 3 integers must be 39. I started at 13 and counted 5 up and 5 down, which already makes sure of the range and the mean is 13 (since it's balanced with 13 being the middle), therefore, 13+5 is 18, 13-5=8.

6 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
2 years ago
A cricket team has a 0.25 chance of losing
Westkost [7]

Answer:

a) 0.125

b) 0.015635

c) 0.00000095367431640625‬

Step-by-step explanation:

a) 0.25^2 = 0.125

b) 0.25^3 = 0.015625

c) 0.25^{10} = 0.00000095367431640625\\\\

8 0
3 years ago
Read 2 more answers
18=3(1+12 x)-5(10x+11)
madreJ [45]
18=3+36x-50x-55

18=3-14x-55

18=-14x-52

70=-14x

x=-5
6 0
3 years ago
Danielle has $6.65 worth of change in nickels and dimes. If she has 5 times as many nickels as dimes, how many of each type of c
evablogger [386]

Answer:

95 nickels 19 dimes

Step-by-step explanation:

d=dimes

5d=nickels

0.05(5d) + 0.10(d) = 6.65

0.25d + 0.10d = 6.65

0.35d=6.65

Divide by 0.35

d= 19 (19 coins)

19 dimes =$1.9

6.65-1.9= $4.75

Nickels=$4.75

4.75/0.05 = 95 coins

19*5=95

  114 coins in total

19*0.10 = 1.9

95*0.05=4.74

4.75+1.9= 6.65

4 0
3 years ago
Read 2 more answers
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