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3 years ago

10

Two twin sisters are away at college. They both called home today. Tara calls home every 4 days. Sara calls home every 3 days. H

ow many days will it be until Tara and Sara both call home on the same day again? (answer only write that number)

1 answer:

frozen [14]3 years ago

7 0

Answer:

12

Step-by-step explanation:

4*3

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3 years ago

Cab 1 charges $1 per kilometer. Cab 2 charges $0.50 per kilometer and a $4 base charge. Solve for the value of k(kilometers ) th

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3 years ago

A student takes a driving test until it is passed.

Leya [2.2K]

Answer:

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

Step-by-step explanation:

The probability that a student passes the driving test at any attempt is,

<em>p</em> = 4/7.

The event of a student passing in any attempt is independent of each other.

The probability that the test is taken an even number of times is:

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

+ P (Passing in the 4th attempt)

+ P (Passing in the 6th attempt) ...

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