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vladimir2022 [97]
3 years ago
10

H(n) = n + 2 g(n) = 2n + 1 Find h(n) − g(n)​

Mathematics
1 answer:
Serggg [28]3 years ago
3 0

Answer:

-n+1

Step-by-step explanation:

h(n)-g(n) =n+2-2n-1 ,collect like terms

n-2n+2-1 = -n+1

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first we need to get rid of the ( ) so that we can see what the question really says

(3x³-5x² - 7+ 4x4) + (9x4 - 10x² - 5x³ + 3)

= 3x³-5x²-7+4x⁴+9x⁴-10x²-5x³+3

now all we need to do is add up the numbers with the same x's

<em>3x³</em>-5x²-7<u>+4x⁴+9x⁴</u>-10x²<em>-5x³</em>+3

= 13x⁴-2x³-15x²-4

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Kim rolls a dice and flips a coin. b) work out the probability that she gets an even number and a tail
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A)3/6=1/2
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3 years ago
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Simplify the following expression:<br><br> (4x+8) - 3(x+1)
Umnica [9.8K]

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x + 5

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  • 4x + 8 + (-3)(x) + (-3)(1)
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Therefore, the answer is x + 5.

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Seventy percent of all vehicles examined at a certain emissions inspection station pass the inspection. Assuming that successive
LenaWriter [7]

Answer:

(a) The probability that all the next three vehicles inspected pass the inspection is 0.343.

(b) The probability that at least 1 of the next three vehicles inspected fail is 0.657.

(c) The probability that exactly 1 of the next three vehicles passes is 0.189.

(d) The probability that at most 1 of the next three vehicles passes is 0.216.

(e) The probability that all 3 vehicle passes given that at least 1 vehicle passes is 0.3525.

Step-by-step explanation:

Let <em>X</em> = number of vehicles that pass the inspection.

The probability of the random variable <em>X</em> is <em>P (X) = 0.70</em>.

(a)

Compute the probability that all the next three vehicles inspected pass the inspection as follows:

P (All 3 vehicles pass) = [P (X)]³

                                    =(0.70)^{3}\\=0.343

Thus, the probability that all the next three vehicles inspected pass the inspection is 0.343.

(b)

Compute the probability that at least 1 of the next three vehicles inspected fail as follows:

P (At least 1 of 3 fails) = 1 - P (All 3 vehicles pass)

                                   =1-0.343\\=0.657

Thus, the probability that at least 1 of the next three vehicles inspected fail is 0.657.

(c)

Compute the probability that exactly 1 of the next three vehicles passes as follows:

P (Exactly one) = P (1st vehicle or 2nd vehicle or 3 vehicle)

                         = P (Only 1st vehicle passes) + P (Only 2nd vehicle passes)

                              + P (Only 3rd vehicle passes)

                       =(0.70\times0.30\times0.30) + (0.30\times0.70\times0.30)+(0.30\times0.30\times0.70)\\=0.189

Thus, the probability that exactly 1 of the next three vehicles passes is 0.189.

(d)

Compute the probability that at most 1 of the next three vehicles passes as follows:

P (At most 1 vehicle passes) = P (Exactly 1 vehicles passes)

                                                       + P (0 vehicles passes)

                                              =0.189+(0.30\times0.30\times0.30)\\=0.216

Thus, the probability that at most 1 of the next three vehicles passes is 0.216.

(e)

Let <em>X</em> = all 3 vehicle passes and <em>Y</em> = at least 1 vehicle passes.

Compute the conditional probability that all 3 vehicle passes given that at least 1 vehicle passes as follows:

P(X|Y)=\frac{P(X\cap Y)}{P(Y)} =\frac{P(X)}{P(Y)} =\frac{(0.70)^{3}}{[1-(0.30)^{3}]} =0.3525

Thus, the probability that all 3 vehicle passes given that at least 1 vehicle passes is 0.3525.

7 0
2 years ago
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