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3 years ago

6

Can someone help me with this question? It's a simple question compared to all the other questions on this app. Many appreciatio

ns!!

1 answer:

IgorLugansk [536]3 years ago

5 0

Do 25×6
then divide that by 160

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What happens if you fail one semester in 8th grade?

Hoochie [10]

Answer:

youll be fine ididnt pass and im now in 10th

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Step-by-step explanation:

7 0

3 years ago

Suppose a random variable x is best described by a uniform probability distribution with range 22 to 55. Find the value of a tha

const2013 [10]

Answer:

(a) The value of <em>a</em> is 53.35.

(b) The value of <em>a</em> is 38.17.

(c) The value of <em>a</em> is 26.95.

(d) The value of <em>a</em> is 25.63.

(e) The value of <em>a</em> is 12.06.

Step-by-step explanation:

The probability density function of <em>X</em> is:

$f_{X}(x)=\frac{1}{55-22}=\frac{1}{33}$

Here, 22 < X < 55.

(a)

Compute the value of <em>a</em> as follows:

$P(X\leq a)=\int\limits^{a}_{22} {\frac{1}{33}} \, dx \\\\0.95=\frac{1}{33}\cdot \int\limits^{a}_{22} {1} \, dx \\\\0.95\times 33=[x]^{a}_{22}\\\\31.35=a-22\\\\a=31.35+22\\\\a=53.35$

Thus, the value of <em>a</em> is 53.35.

(b)

Compute the value of <em>a</em> as follows:

$P(X< a)=\int\limits^{a}_{22} {\frac{1}{33}} \, dx \\\\0.95=\frac{1}{33}\cdot \int\limits^{a}_{22} {1} \, dx \\\\0.49\times 33=[x]^{a}_{22}\\\\16.17=a-22\\\\a=16.17+22\\\\a=38.17$

Thus, the value of <em>a</em> is 38.17.

(c)

Compute the value of <em>a</em> as follows:

$P(X\geq a)=\int\limits^{55}_{a} {\frac{1}{33}} \, dx \\\\0.85=\frac{1}{33}\cdot \int\limits^{55}_{a} {1} \, dx \\\\0.85\times 33=[x]^{55}_{a}\\\\28.05=55-a\\\\a=55-28.05\\\\a=26.95$

Thus, the value of <em>a</em> is 26.95.

(d)

Compute the value of <em>a</em> as follows:

$P(X\geq a)=\int\limits^{55}_{a} {\frac{1}{33}} \, dx \\\\0.89=\frac{1}{33}\cdot \int\limits^{55}_{a} {1} \, dx \\\\0.89\times 33=[x]^{55}_{a}\\\\29.37=55-a\\\\a=55-29.37\\\\a=25.63$

Thus, the value of <em>a</em> is 25.63.

(e)

Compute the value of <em>a</em> as follows:

$P(1.83\leq X\leq a)=\int\limits^{a}_{1.83} {\frac{1}{33}} \, dx \\\\0.31=\frac{1}{33}\cdot \int\limits^{a}_{1.83} {1} \, dx \\\\0.31\times 33=[x]^{a}_{1.83}\\\\10.23=a-1.83\\\\a=10.23+1.83\\\\a=12.06$

Thus, the value of <em>a</em> is 12.06.

7 0

3 years ago

Find f(-2) for the following function. <br><br>f(x)=3.6x-2

butalik [34]

Answer:

f(- 2) = - 9.2

Step-by-step explanation:

To evaluate f(- 2) substitute x = - 2 into f(x) , that is

f(- 2) = 3.6(- 2) - 2 = - 7.2 - 2 = - 9.2

4 0

3 years ago

The graph shows the 2005 median income of some year-round workers and the number of years of school. Estimate the median income

xxTIMURxx [149]

Answer:20??

Step-by-step explanation:

4 0

3 years ago

SOMEONE HELP ME FIND THE Y PLEASE ASAP

Evgesh-ka [11]

Answer:

(Down Below)

Step-by-step explanation:

Plug in the value of x in the left column for all the values in the equation. I will do a couple for you.

Since -4 is the value of x, substitute it into the equation for x.

$y= -4-5=-9$

-1

$y=-1-5=-6$

0

$y=0-5=-5$

You can do the rest. You got this! Comment if you need more help!

3 0

2 years ago

Read 2 more answers