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2 years ago

Combine any like terms in the expression. If there are no like terms, rewrite the expression. 8fg^2 + 7fg^2 – 2fg^2

Mathematics

2 answers:

AleksAgata [21]2 years ago

6 0

13fg^2
Hope this helps!

dlinn [17]2 years ago

5 0

13fg^2 I’m pretty sure this is the answer hope this helps

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The ratio of cats to dogs at an animal rescue is 70:63. If the ratio were to stay the same, how many dogs will there be if there

ratelena [41]

Answer:

There are 9 dogs

Step-by-step explanation:

$\frac{10}{x} = \frac{70}{63} \\ x = \frac{630}{70} \\ x = 9$

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2 years ago

What is the solution to the inequality below? -1x + 5 < 20

Reil [10]

Answer:

x < -15

Step-by-step explanation:

-1x + 5 < 20

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2 years ago

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soldier1979 [14.2K]

9514 1404 393

Answer:

11.8

Step-by-step explanation:

The number has no digits to the right of the tenths place, so the number is correctly rounded as is.

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2 years ago

Solve x^2 - 8x = 3 by completing the square. Which is the solution set of the equation?

sp2606 [1]

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Step-by-step explanation:

I have answered ur question

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2 years ago

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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

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

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

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

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

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2 years ago