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gizmo_the_mogwai [7]

2 years ago

Mathematics

1 answer:

pashok25 [27]2 years ago

8 0

Answer:

Let's call:

f = price of 1 cup of dried fruit

a = price of 1 cup of almonds

In order to build the linear system, you need to consider that the total price of a bag is given by the sum of the price of cups times the number of cups in each bag, therefore:

Solve for a in first equation:

a = (6 - 3f) / 4

Then substitute in the second equation:

41/2 f + 6 · (6 - 3f) / 4 = 9

41/2 f + 9 - 9/2 f = 9

16 f = 0

f = 0

Now, substitute this value in the formula found for a:

a = (6 - 3·0) / 4

= 3/2 = 1.5

Hence, the cups of dried fruit are free and 1 cup of almond costs 1.5$

Step-by-step explanation:

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inn [45]

J:V = 3:2

3:2 x 40 = 120:80

J: 120-15 = 105
V: 80+15 = 95

J:V is now 105:95

The original amount of peas that Julia had at first was 120.

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3 years ago

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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.

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The answer to this problem is 3. 3^7 is equal to 343.

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3 years ago

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Answer:

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Step-by-step explanation:

Look for the sequence in which adjacent terms are related by a common ratio.

A. 10/6 ≠ 6/2

B. 9/5 ≠ 5/3

C. 8/4 ≠ 24/8

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Answer:

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2 years ago

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