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3 years ago

–m + 0.02 + 2.1m = –1.45 – 4.81m 0.01 0.1 10 100

Mathematics

2 answers:

marusya05 [52]3 years ago

8 0

I believe that it is 0.1

Anit [1.1K]3 years ago

3 0

Answer:

D. 100.

Step-by-step explanation:

We have been given an equation $-m+0.02+2.1m=-1.45-4.81m$ . We are supposed to solve our given equation.

To solve our given equation, we will multiply both sides of our given equation by 100 as the term $0.02,-1.45 \text{ and } -4.81$ has two digits after decimal.

Upon multiplying both sides of our given equation by 100, we will get:

$-100m+2+210m=-145-481m$

Therefore, option D is the correct choice.

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Leya [2.2K]

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P (even number of tests) = P (Passing in the 2nd attempt)

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If a student passed in the 2nd attempt it implies that he failed in the first.

Then, P (Passing in the 2nd attempt) = $(\frac{3}{7}) \times (\frac{4}{7})$

Similarly, P (Passing in the 4th attempt) = $(\frac{3}{7})^{3} \times (\frac{4}{7})$ , since he failed in the first 3 attempts.

And so on.

Compute the probability of an even number of tests as follows:

P (even number of tests) = $(\frac{3}{7}) \times (\frac{4}{7})+(\frac{3}{7})^{3} \times (\frac{4}{7})+(\frac{3}{7})^{5} \times (\frac{4}{7})+...$

The result follows a Geometric progression for infinite values.

The sum of infinite GP is:

$S=\frac{a}{1-r^{2}}$

The probability is:

P (even number of tests) = $(\frac{3}{7}) \times (\frac{4}{7})+(\frac{3}{7})^{3} \times (\frac{4}{7})+(\frac{3}{7})^{5} \times (\frac{4}{7})+...$

$=\frac{(\frac{3}{7})(\frac{4}{7} ) }{1-(\frac{3}{7})^{2}}\\=\frac{12}{49}\times\frac{49}{40}\\ =\frac{12}{40}\\ =0.30$

Thus, the probability that the test is taken an even number of times is 0.30.

7 0

3 years ago