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3 years ago

What is the expression in radical form?

Mathematics

1 answer:

RoseWind [281]3 years ago

3 0

Answer:

The answer is;

4^3/10 • x^9/10 •y^3/5

Step-by-step explanation:

We want to express the expression in the bracket in radical form;

(4x^3y^2)^3/10

What we shall do here is to multiply all the powers of the terms in the bracket by 3/10

So we shall have;

4^3/10 • x^(3*3/10) * y^(2*3/10)

= 4^3/10 • x^9/10 • y^3/5

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they are both round, and both go through multiple areas.

Step-by-step explanation:

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Based on the Census, state A has 17 fewer electoral votes for president then state B. If the total number of electoral votes for

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X = B.......x - 17 = A

x + x - 17 = 87
2x - 17 = 87
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3 years ago

Given that Flx) = x2 + 2, evaluate F(2).

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Step-by-step explanation:

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3 years ago

Suppose that 40 percent of the drivers stopped at State Police checkpoints in Storrs on Spring Weekend show evidence of driving

lesantik [10]

Answer:

a) 0.778

b) 0.9222

c) 0.6826

d) 0.3174

e) 2 drivers

Step-by-step explanation:

Given:

Sample size, n = 5

P = 40% = 0.4

a) Probability that none of the drivers shows evidence of intoxication.

$P(x=0) = ^nC_x P^x (1-P)^n^-^x$

$P(x=0) = ^5C_0 (0.4)^0 (1-0.4)^5^-^0$

$P(x=0) = ^5C_0 (0.4)^0 (0.60)^5$

$P(x=0) = 0.778$

b) Probability that at least one of the drivers shows evidence of intoxication would be:

P(X ≥ 1) = 1 - P(X < 1)

$= 1 - P(X = 0)$

$= 1 - ^5C_0 (0.4)^0 * (0.6)^5$

$= 1 - 0.0778$

$= 0.9222$

c) The probability that at most two of the drivers show evidence of intoxication.

P(x≤2) = P(X = 0) + P(X = 1) + P(X = 2)

$^5C_0 (0.4)^0 (0.6)^5 + ^5C_1 (0.4)^1 (0.6)^4 + ^5C_2 (0.4)^2 (0.6)^3$

$= 0.6826$

d) Probability that more than two of the drivers show evidence of intoxication.

P(x>2) = 1 - P(X ≤ 2)

$= 1 - [^5C_0 (0.4)^0 (0.6)^5 + ^5C_1 (0.4)^1 (0.6)^4 + ^5C_2 * (0.4)^2 (0.6)^3]$

$= 1 - 0.6826$

$= 0.3174$

e) Expected number of intoxicated drivers.

To find this, use:

Sample size multiplied by sample proportion

n * p

= 5 * 0.40

= 2

Expected number of intoxicated drivers would be 2

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