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3 years ago

F’(x)= 1+3 square root x f(9)=68

Mathematics

1 answer:

notka56 [123]3 years ago

8 0

Answer:

$f(x) = x + 2x^\frac{3}{2} + 5$

Step-by-step explanation:

I'm assuming the question here is asking for f(x)

The first step would be to find the antiderivative of f'(x)

$\int{1+3\sqrt{x}} \, dx = \int{1 + 3x^\frac{1}{2}} dx$

Apply the reverse power rule

$x + 2x^\frac{3}{2} + C$

Now solve for C when x = 9

$x + 2x^\frac{3}{2} + C = 68\\9 + 2(9)^\frac{3}{2} + C = 68\\9 + 2(27) + C = 68\\9 + 54 + C = 68\\63 + C = 68\\C = 5$

Now substitute into the antiderivative

$f(x) = x + 2x^\frac{3}{2} + 5$

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2 years ago

Simplify completely''''

34kurt

Answer:

$\displaystyle \frac{2(x + 3)}{5x - 2}$

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Step-by-step explanation:

Step 1: Define

$\displaystyle \frac{2x^2 + 16x + 30}{5x^2 + 13x - 6}$

Step 2: Simplify

[Fraction] Factor numerator: $\displaystyle \frac{2(x + 3)(x + 5)}{5x^2 + 13x - 6}$
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2 years ago

For each of the following binomial random variables, specify n and p. (a) A fair die is rolled 50 times. X = number of times a 5

Keith_Richards [23]

Answer:

a) $n = 50, p = \frac{1}{6}$

b) $n = 16, p = \frac{1}{100}$

c) $n = 26, p = 0.25, \mu = 6.5$

Step-by-step explanation:

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinatios of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And $p$ is the probability of X happening.

(a) A fair die is rolled 50 times. X = number of times a 5 is rolled

The die is rolled 50 times, so $n = 50$ .

Each roll can have 6 outcomes. So the probability that 5 is rolled is $p = \frac{1}{6}$

(b) A company puts a game card in each box of cereal and 1/100 of them are winners. You buy sixteen boxes of cereal, and X = number of times you win.

You buy 16 boxes of cereal, so $n = 16$ .