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3 years ago

BRAINLIEST!!! I need help ASAP how do I find the percent change from 1960 to 2000

Mathematics

1 answer:

SVEN [57.7K]3 years ago

4 0

Answer:

percentage change from 1960 to 1970 =( 0.11 ÷0.25 ) × 100

= 44 %

Percentage change from 1970 to 1980 =( 0.83 ÷ 0.36) × 100

= 230.55%

percentage change from 1980 to 1990 = ( 0.16 ÷ 1.19) × 100

= 13.44%

percentage change from 1990 to 2000 = (- 0.09 ÷ 1.35) × 100

= -6.667%

Step-by-step explanation:

Item : 1960 1970 1980 1990 2000

gallon: $0.25 $0.36 $1.19 $1.35 $1.26

change : 0.36 - 0.25 = 0.11

percentage change from 1960 to 1970 =( 0.11 ÷0.25 ) × 100

= 44 %

Percentage change from 1970 to 1980 =( 0.83 ÷ 0.36) × 100

= 230.55%

percentage change from 1980 to 1990 = ( 0.16 ÷ 1.19) × 100

= 13.44%

percentage change from 1990 to 2000 = (- 0.09 ÷ 1.35) × 100

= -6.667%

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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.

= 6

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

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