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3 years ago

There are two formulae you can use to calculate the number of apple trees and the number of conifer

Mathematics

1 answer:

Leno4ka [110]3 years ago

5 0

9514 1404 393

Answer:

n = 8

Step-by-step explanation:

"By inspection" is an appropriate method.

We are asked to compare the expressions

n·n

8·n

and find the value(s) of n that makes them equal. By inspection, we see that n=8 will make these expressions equal. We also know that both expressions will be zero when n=0.

More formally, we could write ...

n^2 = 8n . . . . the two formulas give the same value

n^2 -8n = 0 . . . . rearrange to standard form

n(n -8) = 0 . . . . . factor

Using the zero product rule, we know the solutions will be the values of n that make the factors zero. Those values are ...

n = 0 . . . . . makes the factor n = 0

n = 8 . . . . . makes the factor (n-8) = 0

Generally, we're not interested in "trivial" solutions (n=0), so the only value of n that is of interest is n = 8.

A lot of times, I find a graphing calculator to be a quick and easy way to find function argument values that make expressions equal.

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student randomly receive 1 of 4 versions(A, B, C, D) of a math test. What is the probability that at least 3 of the 5 student te

alexdok [17]

Answer:

1.2%

Step-by-step explanation:

We are given that the students receive different versions of the math namely A, B, C and D.

So, the probability that a student receives version A = $\frac{1}{4}$ .

Thus, the probability that the student does not receive version A = $1-\frac{1}{4}$ = $\frac{3}{4}$ .

So, the possibilities that at-least 3 out of 5 students receive version A are,

1) 3 receives version A and 2 does not receive version A

2) 4 receives version A and 1 does not receive version A

3) All 5 students receive version A

Then the probability that at-least 3 out of 5 students receive version A is given by,

$\frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{3}{4}\times \frac{3}{4}$ + $\frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{3}{4}$ + $\frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}$