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zaharov [31]
3 years ago
7

A home improvement store advertises 60 square feet of flooring for $287.00 including an $80.00 installation fee. How much does e

ach square foot of flooring cost?
Mathematics
1 answer:
Tpy6a [65]3 years ago
6 0
$287-80 for installation leaves $207 for flooring cost
207/60 = $3.44/square foot
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Which expression represents the algebraic phrase “twelve times the sum of a number and seven-tenths”? 12 y + StartFraction 7 ove
kykrilka [37]

Answer:

The answer is:

12 ( y+ \frac{7}{10})

this suits perfectly the description

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What is the answer For -6k Plus 7K
Furkat [3]
-6k+7k equals 1k.................................
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What is 5x15 and 7x6
fiasKO [112]
5x15 is 75 and 7x6 is 42
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Square of a standard normal: Warmup 1.0 point possible (graded, results hidden) What is the mean ????[????2] and variance ??????
LenaWriter [7]

Answer:

E[X^2]= \frac{2!}{2^1 1!}= 1

Var(X^2)= 3-(1)^2 =2

Step-by-step explanation:

For this case we can use the moment generating function for the normal model given by:

\phi(t) = E[e^{tX}]

And this function is very useful when the distribution analyzed have exponentials and we can write the generating moment function can be write like this:

\phi(t) = C \int_{R} e^{tx} e^{-\frac{x^2}{2}} dx = C \int_R e^{-\frac{x^2}{2} +tx} dx = e^{\frac{t^2}{2}} C \int_R e^{-\frac{(x-t)^2}{2}}dx

And we have that the moment generating function can be write like this:

\phi(t) = e^{\frac{t^2}{2}

And we can write this as an infinite series like this:

\phi(t)= 1 +(\frac{t^2}{2})+\frac{1}{2} (\frac{t^2}{2})^2 +....+\frac{1}{k!}(\frac{t^2}{2})^k+ ...

And since this series converges absolutely for all the possible values of tX as converges the series e^2, we can use this to write this expression:

E[e^{tX}]= E[1+ tX +\frac{1}{2} (tX)^2 +....+\frac{1}{n!}(tX)^n +....]

E[e^{tX}]= 1+ E[X]t +\frac{1}{2}E[X^2]t^2 +....+\frac{1}{n1}E[X^n] t^n+...

and we can use the property that the convergent power series can be equal only if they are equal term by term and then we have:

\frac{1}{(2k)!} E[X^{2k}] t^{2k}=\frac{1}{k!} (\frac{t^2}{2})^k =\frac{1}{2^k k!} t^{2k}

And then we have this:

E[X^{2k}]=\frac{(2k)!}{2^k k!}, k=0,1,2,...

And then we can find the E[X^2]

E[X^2]= \frac{2!}{2^1 1!}= 1

And we can find the variance like this :

Var(X^2) = E[X^4]-[E(X^2)]^2

And first we find:

E[X^4]= \frac{4!}{2^2 2!}= 3

And then the variance is given by:

Var(X^2)= 3-(1)^2 =2

7 0
3 years ago
Write the equation of a line in point slope form that is PARALLEL 5x + 2y = 1 that passes through the point (- 2, - 3) .
aliya0001 [1]

Answer:

y = -5/2x - 13

Step-by-step explanation:

First, we have to find the slope of the original equation

Subtract 5x from both sides

5x + 2y = 1

- 5x         - 5x

2y = -5x + 1

Divide both sides by 2

2y/2 = (-5x + 1)/2

y = -5/2x + 1/2

The slope of this equation is -5/2, so -5/2 has to be in our new parallel equation

y = -5/2x + b

Plug in the given x and y

-3 = -5/2(-2) + b

-3 = 10 + b

Subtract 10 from both sides

-3 = 10 + b

- 10  - 10

b = -13

This makes our equation: y = -5/2x - 13

7 0
3 years ago
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