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2 years ago

Find the product.

Mathematics

2 answers:

RUDIKE [14]2 years ago

8 0

Answer:

C. 3x2 − 3x − 60

Step-by-step explanation:

3(x + 4) (x - 5) = (3x+12) (x-5)

(3x+12) (x-5) = 3x2+12x-15x-60

3x2+12x-15x-60 = 3x2 − 3x − 60

LuckyWell [14K]2 years ago

6 0

C. 3x^2-3x-60 Hope it helps

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What is 2 and 5/8 shown as a decimal

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The answer to your question is 2.625

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2 years ago

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What operation does the Product Rule use

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Answer:

addition

Step-by-step explanation:

Example: 5^1 * 5^3= 5^4

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2 years ago

A study showed that 25% of the students drive themselves to school. Based on the suggested probability, in a class of 18 student

Gwar [14]

Answer:

B) 0.283

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 combinations 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.

25% of the students drive themselves to school.

This means that $p = 0.25$

Class of 18 students

This means that $n = 18$

What would be the probability that at least 6 students drive themselves to school?

This is

$P(X \geq 6) = 1 - P(X < 6)$

In which

$P(X < 6) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5)$

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

$P(X = 0) = C_{15,0}.(0.25)^{0}.(0.75)^{18} = 0.006$

$P(X = 1) = C_{15,1}.(0.25)^{1}.(0.75)^{17} = 0.034$

$P(X = 2) = C_{15,2}.(0.25)^{2}.(0.75)^{16} = 0.096$

$P(X = 3) = C_{15,3}.(0.25)^{3}.(0.75)^{15} = 0.17$

$P(X = 4) = C_{15,4}.(0.25)^{4}.(0.75)^{14} = 0.213$

$P(X = 5) = C_{15,5}.(0.25)^{5}.(0.75)^{13} = 0.199$

$P(X < 6) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) = 0.006 + 0.034 + 0.096 + 0.17 + 0.213 + 0.199 = 0.718$

$P(X \geq 6) = 1 - P(X < 6) = 1 - 0.718 = 0.282$

Closest option is B, just a small rounding difference.

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2 years ago

How to solve quadratic equations

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If we have two numbers A&B and we know that A times B=0 then it must follow that either A=0, or B=0 (or both)

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2 years ago

Find dy/dx of the function y = √x sec*-1 (√x)

ELEN [110]

Hi there!

$\large\boxed{\frac{dy}{dx} = \frac{1}{2\sqrt{x}}sec^{-1}(\sqrt{x}) + \frac{1}{2|\sqrt{x}|\sqrt{{x} - 1}}}$

$y = \sqrt{x} * sec^{-1}(-\sqrt{x}})$

Use the chain rule and multiplication rules to solve:

g(x) * f(x) = f'(x)g(x) + g'(x)f(x)

g(f(x)) = g'(f(x)) * 'f(x))

Thus:

f(x) = √x

g(x) = sec⁻¹ (√x)

$\frac{dy}{dx} = \frac{1}{2\sqrt{x}}sec^{-1}(\sqrt{x}) + \sqrt{x} * \frac{1}{\sqrt{x}\sqrt{\sqrt{x}^{2} - 1}} * \frac{1}{2\sqrt{x}}$

Simplify:

$\frac{dy}{dx} = \frac{1}{2\sqrt{x}}sec^{-1}(\sqrt{x}) + \sqrt{x} * \frac{1}{2|x|\sqrt{{x} - 1}}$

$\frac{dy}{dx} = \frac{1}{2\sqrt{x}}sec^{-1}(\sqrt{x}) + \frac{1}{2|\sqrt{x}|\sqrt{{x} - 1}}$

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