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3 years ago

(4^3) 2 What is the value of the expression when simplified

Mathematics

2 answers:

Anna007 [38]3 years ago

8 0

Answer:

Hey, Zachary. Your answer would be 128

Step-by-step explanation:

4^3 is 64, and 64 x 2 is 128!

Luden [163]3 years ago

6 0

Answer:

[See Below]

Step-by-step explanation:

$Hey~There!$

_______________________

→ Simplify $(4^3)*2$ :

$4^3=64$

→ Rewrite the new equation:

$64*2$

→ Solve:

$64*2=128$

_______________________

So your answer would be $128$ .

_______________________

$Hope~this~helps!$

$-Shane,~SpamIsTheMan :)$

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3 years ago

Read 2 more answers

Suppose that 10% of all homeowners in an earthquake-prone area of California are insured against earthquake damage. Four homeown

guapka [62]

Remainder of question:

Find the probability distribution of x

Answer:

The random variable x is defined as: X = {0, 1, 2, 3, 4}

The probability distribution of X:

P(X = 0) = 0.656

P(X = 1) = 0.2916

P(X= 2) = 0.0486

P(X=3) = 0.0036

P(X = 4) = 0.0001

Step-by-step explanation:

Sample size, n = 4

Random variable, X = {0, 1, 2, 3, 4}

10% (0.1) of the homeowners are insured against earthquake, p = 0.1

Proportion of homeowners who are not insured against earthquake, q = 1 - 0.1

q = 0.9

Probability distribution of x,

$P(X = r) = ^nC_r *p^r q^{n-r} \\\\P(X= 0) =(^4C_0 *p^1 q^4 )\\P(X=0) = (^4C_0 *0.1^0 0.9^4 ) = 0.656\\P(X= 1)= (^4C_1 *p^1 q^3 )\\P(X=1) = (^4C_1 *0.1^1 0.9^3 ) = 0.2916\\P(X= 2)=( ^4C_2 *p^2 q^2) \\P(X=2) = (^4C_2 *0.1^2 0.9^2 ) = 0.0486\\P(X= 3) = (^4C_3 *p^3 q^3) \\ P(X=3) = (^4C_3 *0.1^3 0.9^1 ) = 0.0036\\P(X= 4) = (^4C_4 *p^4 q^0 )\\ P(X=4) =(^4C_4 *0.1^4 0.9^0 ) = 0.0001$