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3 years ago

6

#3 - Simplify the expression. Type your answer with all lowercase letters and no spaces. Type your p variable first and g variab

le second in your answer. * 2(5p+2g)+5g

1 answer:

Basile [38]3 years ago

7 0

Answer: 10p+9g

Step-by-step explanation:

2(5p+2g)+5g

10p+4g+5g

10p+9g

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He perimeter of the figure is given. Find the length of the indicated side.

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Use the graph to write a linear function that relates y to x

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2 years ago

The probability distribution of the amount of memory X (GB) in a purchased flash drive is given below. x 1 2 4 8 16 p(x) .05 .10

Rama09 [41]

Answer:

a) $E(X) = 6.45$

b) $E(X^{2} )= 57.25$

c) $V(X) = 15.648$

d) E(3X + 2) = 21.35

e) $E(3X^{2} +2) = 173.75$

f) V(3X+2) = 140.832

g) E(X+1) = 7.45

h) V(X+1) = 15.648

Step-by-step explanation:

a) $E(X) = \sum xP(x)$

$E(X) = (1*0.05) + (2*0.10) + (4*0.35) + (8*0.40) + (16*0.10)\\E(X) = 6.45$

b)

$E(X^{2} ) = (1^{2} *0.05) + (2^{2} *0.10) + (4^{2} *0.35) + (8^{2} *0.40) + (16^{2} *0.10)\\ E(X^{2} )= 57.25$

c)

$V(X) = E(X^{2} ) - (E(X))^{2} \\V(X) = 57.25 - 6.45^{2} \\V(X) = 15.648$

d)

$E(3X+2) = 3E(X) + 2\\E(3X+2) = (3*6.45) + 2 \\E(3X+2) = 21.35$

e)

$E(3X^{2} +2) = 3E(X^{2} ) + 2\\E(3X^{2} +2) = (3*57.25) + 2 \\E(3X^{2} +2) = 173.75$

f)

$V(3X+2) = 3^{2} V(X)\\V(3X+2) = 9*15.648\\V(3X+2) = 140.832$

g)

$E(X+1) = E(X) + 1\\E(X+1) = 6.45 + 1\\E(X+1) =7.45$

h)

$V(X+1) = 1^{2} V(X)\\V(X+1) = 15.648$

6 0

3 years ago

What is the volume of a cone with a diameter of 10 in and a height of 12 in

Radda [10]

Answer:

V = 100π

Step-by-step explanation:

The volume of a cone is V = ( $\frac{\pi r^2h}{3}$ )

1) $\frac{\pi5^212}{3}$

2) $\frac{300\pi }{3}$

3) 100π

5 0

3 years ago

30 POINTS! Consider the system of linear equations and the matrix equations below.

fiasKO [112]

Answer:

Option C is correct.

Step-by-step explanation:

-x+3y=2

4x-2y=22

In matrix form is represented as:

$\left[\begin{array}{cc}-1&3\\4&-2\end{array}\right] \left[\begin{array}{c}x&y\end{array}\right] =\left[\begin{array}{c}2&22\end{array}\right]$

AX=B

$X = A^{-1}B$

$A^{-1} = |A|/Adj A$

|A| = (-1)(-2)-(3)(4)

|A| = 2-12

|A| = -10

Adj A = $\left[\begin{array}{cc}-2&-3\\-4&-1\end{array}\right]$

A^-1 = -1/10 $\left[\begin{array}{cc}-2&-3\\-4&-1\end{array}\right]$

A^-1 = 1/10 $\left[\begin{array}{cc}2&3\\4&1\end{array}\right]$

X= A^-1 B

X = 1/10 $\left[\begin{array}{cc}2&3\\4&1\end{array}\right]$ $\left[\begin{array}{c}2&22\end{array}\right]$

X=1/10 $X=1/10\left[\begin{array}{c}2*2+3*22\\4*2+1*22\end{array}\right]\\X=1/10\left[\begin{array}{c}4+66\\8+22\end{array}\right]\\X=1/10\left[\begin{array}{c}70\\30\end{array}\right]\\X=\left[\begin{array}{c}70/10\\30/10\end{array}\right]\\X=\left[\begin{array}{c}7\\3\end{array}\right]$

So, x = 7 and y =3

Hence Option C is correct.

4 0

3 years ago

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