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3 years ago

15

A number is randomly selected from the set {1,2,3,4,5,6,8}. If the number chosen is greater than 3, what is the probability it i

s odd?

1 answer:

algol [13]3 years ago

5 0

Answer:4

Step-by-step explanation:

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Jackrabbits are capable of reaching speeds up to 40 miles per hour. How fast is this in feet per second? (Round to the nearest w

pishuonlain [190]

Answer:

59 feet per second

Step-by-step explanation:

(40m)(5280ft)=211,200 ft/hr
(211,200 ft/hr )=(211,200ft/3600s)
58.7 ft/s

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2 years ago

Read 2 more answers

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)

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

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