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3 years ago

Let a= x2 +4 rewrite the following equation in terms of a and set it equal to zero. (X2 + 4)2 + 32 = 12x2 + 48 in the resulting

equation what is the coefficient of the a term and what is the constant

Mathematics

1 answer:

Alex73 [517]3 years ago

5 0

$Answer: \\ {( {x}^{2} + 4) }^{2} + 32 = 12 {x}^{2} + 48 \\ \Leftrightarrow {a}^{2} + 32 = 12( {x}^{2} + 4) \\ \Leftrightarrow {a}^{2} + 32 = 12a \\ \Leftrightarrow {a}^{2} - 12a + 32 = 0 \\ \Rightarrow Coefficients:1, \: -12. \: Constant:a$

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3 years ago

How many solutions does the equation 3(x-5)-7=-3x8 have

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Answer:

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Step-by-step explanation:

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3 years ago

At Munder Difflin Paper Company, the manager Mitchell Short randomly places golden sheets of paper inside of 30% of their paper

Korvikt [17]

Answer:

90.67% probability that John finds less than 7 golden sheets of paper

Step-by-step explanation:

For each container, there are only two possible outcomes. Either it contains a golden sheet of paper, or it does not. The probability of a container containing a golden sheet of paper is independent of other containers. So we use the binomial probability distribution to solve this question.

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.

At Munder Difflin Paper Company, the manager Mitchell Short randomly places golden sheets of paper inside of 30% of their paper containers.

This means that $p = 0.3$

14 of these containers of paper.

This means that $n = 14$

What is the probability that John finds less than 7 golden sheets of paper?

$P(X < 7) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6)$

In which

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

$P(X = 0) = C_{14,0}.(0.3)^{0}.(0.7)^{14} = 0.0068$

$P(X = 1) = C_{14,1}.(0.3)^{1}.(0.7)^{13} = 0.0407$

$P(X = 2) = C_{14,2}.(0.3)^{2}.(0.7)^{12} = 0.1134$

$P(X = 3) = C_{14,3}.(0.3)^{3}.(0.7)^{11} = 0.1943$

$P(X = 4) = C_{14,4}.(0.3)^{4}.(0.7)^{10} = 0.2290$

$P(X = 5) = C_{14,5}.(0.3)^{5}.(0.7)^{9} = 0.1963$

$P(X = 6) = C_{14,6}.(0.3)^{6}.(0.7)^{8} = 0.1262$

$P(X < 7) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) = 0.0068 + 0.0407 + 0.1134 + 0.1943 + 0.2290 + 0.1963 + 0.1262 = 0.9067$

90.67% probability that John finds less than 7 golden sheets of paper

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3 years ago

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Mumz [18]

Answer:

R = 8,228,571.43 (Approx)

Step-by-step explanation:

Given:

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x = 4.8 x 10⁵

y = 2.8 x 10⁴

Find:

Value of R

Computation:

R = x²/y

R = (4.8 x 10⁵)²/ (2.8 x 10⁴)

R = 23.04 x 10¹⁰ / (2.8 x 10⁴)

R = 8.22857.. x 10⁶

R = 8,228,571.43 (Approx)

4 0

3 years ago