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3 years ago

10

On a math test, Jessica answered 20% questions incorrectly. She answered 16 questions correctly. How many total questions did sh

e answer?

2 answers:

Tanzania [10]3 years ago

8 0

The answer to the question is 36

shutvik [7]3 years ago

3 0

Answer:

20

Step-by-step explanation:

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The second term of an arithmetic sequence is -39. The rule a, = a -1 + 12 can be used to find the next term of the

siniylev [52]

Step-by-step explanation:

the first one is the correct one

a, = -51+12(n-1)

a(2) = -51+12(2-1)=-39

I hope this helps

Im sure from the answer

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4 0

3 years ago

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A used book store will trade 2 of its books for 3 of yours. If Val brings in 18 books to trade, how many books can she get from

Shalnov [3]

You need to set it up as a fraction and make the fractions equal each other.
2/3 (2 of their books for 3 of yours)
x/18 (x amount of their books for 18 of yours)
(2/3)=(x/18)
multiply both sides by 18
and you get
36/3 which reduces to 12

so your answer is 12

5 0

3 years ago

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What is the unit rate of (32-ounce) bottle of milk cost $1.19

Sholpan [36]

29.87 should be it hopefully

7 0

3 years ago

Ruben bought 6 comic books for $21. Each comic book was the same price.

kondor19780726 [428]

Answer:

$3.50

Step-by-step explanation:

Divide the price of 6 books ($21) by the number of books bought (6)

$21 divided by 6 = $3.50

So the price of 1 book is $3.50

Hope this helps :)

8 0

2 years ago

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The random variable X has the following probability density function: fX(x) = ( xe−x , if x > 0 0, otherwise. (a) Find the mo

dusya [7]

Answer:

Follows are the solution to the given points:

Step-by-step explanation:

Given value:

$\to f_X (x) \ \ xe^{-x} \ , \ x>0$

For point a:

Moment generating function of X=?

Using formula:

$\to M(t) =E(e^{tx})= \int^{\infty}_{-\infty} \ e^{tx} f(x) \ dx$

$M(t) = \int^{\infty}_{-\infty} \ e^{tx}xe^{-x} \ dx = \int^{\infty}_{0} \ xe^{(t-1)x} \ dx$

integrating the values by parts:

$u = x \\\\dv = e^{(t-1)x}\\\\dx =dx \\\\v= \frac{e^{(t-1)x}}{t-1}\\\\M(t) =[\frac{e^{(t-1)x}}{t-1}]^{-\infty}_{0} -\int^{\infty}_{0} \frac{e^{(t-1)x}}{t-1} \ dx\\\\$

$= \frac{1}{t-1} (0) - [\frac{e^{(t-1)x}}{(t-1)^2}]^{\infty}_{0}\\\\=\frac{1}{(t-1)^2}(0-1)\\\\=\frac{1}{(t-1)^2}\\\\$

Therefore, the moment value generating by the function is $=\frac{1}{(t-1)^2}$

In point b:

$E(X^n)=?$

Using formula: $E(X^n)= M^{n}_{X}(0)$

form point (a):

$\to M_{X}(t)=\frac{1}{(t-1)^2}$

Differentiating the value with respect of t

$M'_{X}(t)=\frac{-2}{(t-1)^3}$

when $t=0$

$M'_{X}(0)=\frac{-2}{(0-1)^3}= \frac{-2}{(-1)^3}= \frac{-2}{-1}=2\\\\M''_{X}(t)=\frac{(-2)(-3)}{(t-1)^4}\\\\M''_{X}(0)=\frac{(-2)(-3)}{(0-1)^4}= \frac{6}{(-1)^4}= \frac{6}{1}=6\\\\M''_{X}(t)=\frac{(-2)(-3)}{(t-1)^4}\\\\M''_{X}(0)=\frac{(-2)(-3)}{(0-1)^4}= \frac{6}{(-1)^4}= \frac{6}{1}=6\\\\M'''_{X}(t)=\frac{(-2)(-3)(-4)}{(t-1)^5}\\\\M''_{X}(0)=\frac{(-2)(-3)(-4)}{(0-1)^5}= \frac{24}{(-1)^5}= \frac{24}{-1}=-24\\\\\therefore \\\\M^{K}_{X} (t)=\frac{(-2)(-3)(-4).....(k+1)}{(t-1)^{k+2}}\\\\$

$M^{K}_{X} (0)=\frac{(-2)(-3)(-4).....(k+1)}{(-1)^{k+2}}\\\\\therefore\\\\E(X^n) = \frac{(-2)(-3)(-4).....(n+1)}{(-1)^{n+2}}\\\\$

7 0

2 years ago