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3 years ago

12

Can you please show your work and explain if possible.

1 answer:

Ksju [112]3 years ago

4 0

Answer:

CHEATING ON CLASSWORK

Step-by-step explanation:

IS FORBIDDEN

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Rapido porfa es para hoy.<br> Convierta 0.80 Megagramos a picogramos.

gtnhenbr [62]

Answer:

8 * 10^17 Picogramos

Step-by-step explanation;

El 17 es un exponente. Creo que debes de tartar de tener la tabla de conversion y se te hara mas facil :)

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2 years ago

Find the length of the hypotenuse of a 45°-45°-90° triangle with a leg<br> length of 8 centimeters.

MArishka [77]

Answer:

8 $\sqrt{2$ = $\sqrt{128$ = 11.31

Step-by-step explanation:

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3 years ago

When using a number line to model addition, what does the first integer in a subtraction problem tell you?

Sindrei [870]

You answers your own question

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2 years ago

Which two quadrants have negative x-values?

mart [117]

Answer:

II and III ( 2 and 3)

Step-by-step explanation:

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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