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2 years ago

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A total of 348 teachers are attending a conference. Suppose 195 of those attending are math teachers, and the other attendees ar

e science teachers (x). Which THREE statements are correct for this situation?
(I need help quick)

1 answer:

umka2103 [35]2 years ago

5 0

Answer choice is b c d

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

polet [3.4K]

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).

= 0.8929

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

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

5 0

3 years ago

Help plzzzzzzzzzzzzz

ZanzabumX [31]

1 1/2 * 3/4 =

Change 1 1/2 to an improper fraction = 3/2

3/2 * 3/4 = 6/4 * 3/4

6/4 * 3/4 = 18/4 = 4 2/4 = 4 1/2

The answer is 4 1/2.

Hope this helped☺☺

4 0

3 years ago

Suppose you spend $23, and you pay with a $50 bill. The cashier gives your change in bills. List all the different ways you can

Aleksandr [31]

Here are some answer i can think of: 2-10$ 1-5$ 2-1$, 6-5$ 2-1$, 27-1$, 1-10$ 3-5$ 2-1$

3 0

3 years ago

If a-b/b = 3/7 What is b/a

Basile [38]

Answer:

7/10

Step-by-step explanation:

Solve for a by simplifying both sides of the equation, then isolating the variable.

Glad I could help!

6 0

3 years ago

Fine the sum of the series

lapo4ka [179]

Answer:

6825

Step-by-step explanation:

From the equation, the series is a geometric series, such that

a = 5

r = 4

n = 6

sum of a geometric series is a(r^n - 1)/(r - 1)

therefore,

sum of the series is 5 x (4^6 - 1) / (4 - 1)

= 5 x (4096 - 1) / 3

= 5 x 4095 / 3

= 6825

6 0

3 years ago