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Burka [1]
3 years ago
5

Mia is three years older than twice her sister brooke's age. The sum of their ages is less than 30. what is the greatest age Bro

oke can be?
Mathematics
1 answer:
Pachacha [2.7K]3 years ago
7 0

Answer: 9

Step-by-step explanation: You can start by setting this up as an equation, where X is Brooke’s age.

X + 2(x)+ 3 = 30

The 2(X) + 3 signifies the part of the problems identifying that Mia is twice her sisters age plus 3 years.

Combine x’s

3x + 3 = 30

Subtract 3 from both sides to isolate the x term

3x= 27

Divide both sides by 3 to reduce to a single x

X= 9 , which is brooke’s age

You might be interested in
The probability that your call to a service line is answered in less than 30 seconds is 0.75. Assume that your calls are indepen
vfiekz [6]

Answer:

a) 0.2581

b) 0.4148

c) 17

Step-by-step explanation:

For each call, there are only two possible outcomes. Either they are answered in less than 30 seconds. Or they are not. The probabilities for each call are independent. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

In which C_{n,x} is the number of different combinations of x objects from a set of n elements, given by the following formula.

C_{n,x} = \frac{n!}{x!(n-x)!}

And p is the probability of X happening.

In this problem we have that:

p = 0.75

a. If you call 12 times, what is the probability that exactly 9 of your calls are answered within 30 seconds? Round your answer to four decimal places (e.g. 98.7654).

This is P(X = 9) when n = 12. So

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 9) = C_{12,9}.(0.75)^{9}.(0.25)^{3} = 0.2581

b. If you call 20 times, what is the probability that at least 16 calls are answered in less than 30 seconds? Round your answer to four decimal places (e.g. 98.7654).

This is P(X \geq 16) when n = 20

So

P(X \geq 16) = P(X = 16) + P(X = 17) + P(X = 18) + P(X = 19) + P(X = 20)

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 16) = C_{20,16}.(0.75)^{16}.(0.25)^{4} = 0.1897

P(X = 17) = C_{20,17}.(0.75)^{17}.(0.25)^{3} = 0.1339

P(X = 18) = C_{20,18}.(0.75)^{18}.(0.25)^{2} = 0.0669

P(X = 19) = C_{20,19}.(0.75)^{19}.(0.25)^{1} = 0.0211

P(X = 20) = C_{20,20}.(0.75)^{20}.(0.25)^{0} = 0.0032

So

P(X \geq 16) = P(X = 16) + P(X = 17) + P(X = 18) + P(X = 19) + P(X = 20) = 0.1897 + 0.1339 + 0.0669 + 0.0211 + 0.0032 = 0.4148

c. If you call 22 times, what is the mean number of calls that are answered in less than 30 seconds? Round your answer to the nearest integer.

The expected value of the binomial distribution is:

E(X) = np

In this question, we have n = 22

So

E(X) = 22*0.75 = 16.5

The closest integer to 16.5 is 17.

7 0
3 years ago
Which choice is the equation of a line that passes through (9, -3) and is parallel to the line represented by this equation? y=5
miskamm [114]

Answer:

y+3=\frac{5}{3}(x-9)  or  y=\frac{5}{3}x-18

Step-by-step explanation:

step 1

Find the slope of the line parallel to the given line

we know that

If two lines are parallel, then their slopes are the same

we have

y=\frac{5}{3}x-4

therefore

The slope m of the line parallel to the given line is

m=\frac{5}{3}

step 2

Find the equation of the line in point slope form

y-y1=m(x-x1)

we have

m=\frac{5}{3}

point\ (9,-3)

substitute

y+3=\frac{5}{3}(x-9) ----> equation of the line in point slope form

step 3

Find the equation of the line in slope intercept form

y=mx+b

we have

y+3=\frac{5}{3}(x-9)

Isolate the variable y

y+3=\frac{5}{3}x-15

y=\frac{5}{3}x-15-3

y=\frac{5}{3}x-18

7 0
3 years ago
Find the surface area of the pyramid.
Sphinxa [80]

SA= a+1/2(p)(h)

SA= 18+1/2( 18)(5.2)

SA= 18+46.8

SA=64.8

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3 years ago
Will 3600 to 3200 be and increase or decrease and what is the percentage
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Decrease 11.11%

I know this I did this on brainly
5 0
3 years ago
Read 2 more answers
PLEASE HELP!!
Nadusha1986 [10]
M(meters) = 0.9144*y(yards)
4 0
3 years ago
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