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djverab [1.8K]
3 years ago
9

Choose all the numbers that are common multiples of 3 and 12

Mathematics
1 answer:
Andre45 [30]3 years ago
6 0

Answer:

Step-by-step explanation:

-48

-72

-84

basically if its divisible by 12 its divisible by 3

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What number should both sides of the following equation be divided by to solve for x?
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5. Which expression has the value of -2?
julia-pushkina [17]
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Find the percent of decrease from 340 to 160. Round to the nearest tenth of a
PIT_PIT [208]

Answer: D is correct, 52.9%

Step-by-step explanation:

% decrease from 340 to 160, Where 340 is the old number and 160 is the new number

Percent change = (New number - Old number) / Old number x 100%

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= (160 - 340) / 340 x 100

= (-180) / 340x 100 = -52.94%

I hope this is clear, please mark as brainliest answer.

5 0
4 years ago
The probability distribution of the number of students absent on Mondays, is as follows: X 0 1 2 3 4 5 6 7 f(x) 0.02 0.03 0.26 0
alexgriva [62]

a) Add up all the probabilities f(x) where x>3:

f(4)+f(5)+f(6)+f(7)=0.35

b) The expected value is

E[X]=\displaystyle\sum_xx\,f(x)=3.16

Since X is the number of absent students on Monday, the expectation E[X] is the number of students you can expect to be absent on average on any given Monday. According to the distribution, you can expect around 3 students to be consistently absent.

c) The variance is

V[X]=E[(X-E[X])^2]=E[X^2]-E[X]^2

where

E[X^2]=\displaystyle\sum_xx^2\,f(x)=11.58

So the variance is

V[X]=11.58-3.16^2\approx1.59

The standard deviation is the square root of the variance:

\sqrt{V[X]}\approx1.26

d) Since Y=7X+3 is a linear combination of X, computing the expectation and variance of Y is easy:

E[Y]=E[7X+3]=7E[X]+3=25.12

V[Y]=V[7X+3]=7^2V[X]\approx78.13

e) The covariance of X and Y is

\mathrm{Cov}[X,Y]=E[(X-E[X])(Y-E[Y])]=E[XY]-E[X]E[Y]

We have

XY=X(7X+3)=7X^2+3X

so

E[XY]=E[7X^2+3X]=7E[X^2]+3E[X]=90.54

Then the covariance is

\mathrm{Cov}[X,Y]=90.54-3.16\cdot25.12\approx11.16

f) Dividing the covariance by the variance of X gives

\dfrac{\mathrm{Cov}[X,Y]}{V[X]}\approx\dfrac{11.16}{1.59}\approx0.9638

5 0
3 years ago
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