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VLD [36.1K]
3 years ago
14

There are 4 truck in the fleet of 10 cars. 2 cars were randomly selected. Make a series of distribution of a discrete random var

iable X - the number of trucks among the selected.
Mathematics
1 answer:
timofeeve [1]3 years ago
5 0

Answer:

X = 0 , P(0) = 1/3

X = 1 , P(1) = 8/15

X = 2 , P(2) = 2/15

Step-by-step explanation:

Let X be the no. of trucks

X = 0 , P(0) = 6/10 × 5/9 = 1/3

X = 1 , P(1) = 2(6/10 × 4/9) = 8/15

X = 2 , P(2) = 4/10 × 3/9 = 2/15

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55%of92 in percent time
solmaris [256]
That would be <span>50.6, I believe. You multiply 92 by 55% or .55 and get 50.6 </span>
6 0
3 years ago
Expressions and Equations
Anna71 [15]

Answer:

A) 4x+8

B) 8x-16

C) 8x+12

D) 6x+6y+6z

Step-by-step explanation:

multiple the number outside of the parentheses by each number or letter inside the parentheses

8 0
2 years ago
An investigator wants to assess whether the mean m = the weight of passengers flying on small planes exceeds the FAA guideline o
stira [4]

Answer:

H_{0}: \mu= 185 and H_{a}: \mu > 185

Step-by-step explanation:

The null hypothesis H_{0} states that a population parameter (such as the mean, the standard deviation, and so on) is equal to a hypothesized value. We can write the null hypothesis in the form H_{0}: parameter = value

In this context, the investigator's null hypothesis should be that the average total weight is no different than the reported value by the FAA. We can write it in this form H_{0}: \mu= 185.

The alternative hypothesis H_{a} states that a population parameter is smaller, greater, or different than the hypothesized value in the null hypothesis. We can write the alternative hypothesis in one of three forms

H_{a}: parameter > value\\H_{a}: parameter < value\\H_{a}: parameter \neq value

The investigator wants to know if the average weight of passengers flying on small planes exceeds the FAA guideline of the average total weight of 185 pounds. He should use H_{a}: \mu > 185 as his alternative hypothesis.

7 0
3 years ago
Jose asks his friends to guess the higher of two grades he received on his math tests. He gives them two hints. The difference o
tatyana61 [14]

Answer : 96

x – y = 16 --------> equation 1

1/8 x + 1/2 y = 52

x is the higher grade and y is the lower grade

We solve the first equation for y

x - y = 16

-y = 16 -x ( divide each term by -1)

y = -16 + x

Now substitute y in second equation

1/8 x + 1/2 ( -16 + x ) = 52

1/8x - 8 + 1/2 x = 52

1/8x + 1/2x - 8 = 52

Take common denominator to combine fractions

1/8x + 4/8x -8 = 52

5/8x - 8 = 52

Add 8 on both sides

5/8x = 60

Multiply both sides by 8/5

x = 96

We know x is the higher grade

96 is the higher grade of Jose’s two tests.

8 0
3 years ago
Can I get help with finding the Fourier cosine series of F(x) = x - x^2
trapecia [35]
Assuming you want the cosine series expansion over an arbitrary symmetric interval [-L,L], L\neq0, the cosine series is given by

f_C(x)=\dfrac{a_0}2+\displaystyle\sum_{n\ge1}a_n\cos nx

You have

a_0=\displaystyle\frac1L\int_{-L}^Lf(x)\,\mathrm dx
a_0=\dfrac1L\left(\dfrac{x^2}2-\dfrac{x^3}3\right)\bigg|_{x=-L}^{x=L}
a_0=\dfrac1L\left(\left(\dfrac{L^2}2-\dfrac{L^3}3\right)-\left(\dfrac{(-L)^2}2-\dfrac{(-L)^3}3\right)\right)
a_0=-\dfrac{2L^2}3

a_n=\displaystyle\frac1L\int_{-L}^Lf(x)\cos nx\,\mathrm dx

Two successive rounds of integration by parts (I leave the details to you) gives an antiderivative of

\displaystyle\int(x-x^2)\cos nx\,\mathrm dx=\frac{(1-2x)\cos nx}{n^2}-\dfrac{(2+n^2x-n^2x^2)\sin nx}{n^3}

and so

a_n=-\dfrac{4L\cos nL}{n^2}+\dfrac{(4-2n^2L^2)\sin nL}{n^3}

So the cosine series for f(x) periodic over an interval [-L,L] is

f_C(x)=-\dfrac{L^2}3+\displaystyle\sum_{n\ge1}\left(-\dfrac{4L\cos nL}{n^2L}+\dfrac{(4-2n^2L^2)\sin nL}{n^3L}\right)\cos nx
4 0
3 years ago
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