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andre [41]
3 years ago
9

If Josh can read 18 pages in 40 minutes, how long will it take him to read 90 pages?

Mathematics
2 answers:
Dvinal [7]3 years ago
3 0

Answer:

18=40

90? 40×90=3600÷18=200 pages.

Andrei [34K]3 years ago
3 0

Answer:

It will take him 200 minutes to read 90 pages.

Step-by-step explanation:

Josh can read 18 pages in 40 minutes.

Ratio of pages to minutes = 18 : 40

Number of pages to read = 90

Minutes = x

Ratio of pages to minutes = 90 : x

The relationship is proportional

18 : 40 :: 90 : x

Product of extremes = Product of means

18x = 40*90

18x = 3600

Dividing both sides by 18

\frac{18x}{18}=\frac{3600}{18}

x = 200

Therefore,

It will take him 200 minutes to read 90 pages.

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A university found that of its students withdraw without completing the introductory statistics course. Assume that students reg
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Answer:

A university found that 30% of its students withdraw without completing the introductory statistics course. Assume that 20 students registered for the course.

a. Compute the probability that 2 or fewer will withdraw (to 4 decimals).

= 0.0355

b. Compute the probability that exactly 4 will withdraw (to 4 decimals).

= 0.1304

c. Compute the probability that more than 3 will withdraw (to 4 decimals).

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d. Compute the expected number of withdrawals.

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Step-by-step explanation:

This is a binomial problem and the formula for binomial is:

P(X = x) = nCx p^{x} q^{n - x}

a) Compute the probability that 2 or fewer will withdraw

First we need to determine, given 2 students from the 20. Which is the probability of those 2 to withdraw and all others to complete the course. This is given by:

P(X = x) = nCx p^{x} q^{n - x}\\P(X = 2) = 20C2(0.3)^2(0.7)^{18}\\P(X = 2) =190 * 0.09 * 0.001628413597\\P(X = 2) = 0.027845872524

P(X = x) = nCx p^{x} q^{n - x}\\P(X = 1) = 20C1(0.3)^1(0.7)^{19}\\P(X = 1) =20 * 0.3 * 0.001139889518\\P(X = 1) = 0.006839337111

P(X = x) = nCx p^{x} q^{n - x}\\P(X = 0) = 20C0(0.3)^0(0.7)^{20}\\P(X = 0) =1 * 1 * 0.000797922662\\P(X = 0) = 0.000797922662

Finally, the probability that 2 or fewer students will withdraw is

P(X = 2) + P(X = 1) + P(X = 0) \\= 0.027845872524 + 0.006839337111 + 0.000797922662\\= 0.035483132297\\= 0.0355

b) Compute the probability that exactly 4 will withdraw.

P(X = x) = nCx p^{x} q^{n - x}\\P(X = 4) = 20C4(0.3)^4(0.7)^{16}\\P(X = 4) = 4845 * 0.0081 * 0.003323293056\\P(X = 4) = 0.130420974373\\P(X = 4) = 0.1304

c) Compute the probability that more than 3 will withdraw

First we will compute the probability that exactly 3 students withdraw, which is given by

P(X = x) = nCx p^{x} q^{n - x}\\P(X = 3) = 20C3(0.3)^3(0.7)^{17}\\P(X = 3) = 1140 * 0.027 * 0.002326305139\\P(X = 3) = 0.071603672205\\P(X = 3) = 0.0716

Then, using a) we have that the probability that 3 or fewer students withdraw is 0.0355+0.0716=0.1071. Therefore the probability that more than 3 will withdraw is 1 - 0.1071=0.8929

d) Compute the expected number of withdrawals.

E(X) = 3/10 * 20 = 6

Expected number of withdrawals is the 30% of 20 which is 6.

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