93. Let f(x) = 3x - 1 and g(x) = x + 5. Evaluate the following limits. (a) lim x→3 f(g(x)] (b) lim x→3 g[f(x)]

At Munder Difflin Paper Company, the manager Mitchell Short randomly places golden sheets of paper inside of 30% of their paper

Korvikt [17]

Answer:

90.67% probability that John finds less than 7 golden sheets of paper

Step-by-step explanation:

For each container, there are only two possible outcomes. Either it contains a golden sheet of paper, or it does not. The probability of a container containing a golden sheet of paper is independent of other containers. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

At Munder Difflin Paper Company, the manager Mitchell Short randomly places golden sheets of paper inside of 30% of their paper containers.

This means that $p = 0.3$

14 of these containers of paper.

This means that $n = 14$

What is the probability that John finds less than 7 golden sheets of paper?

$P(X < 7) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6)$

In which

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{14,0}.(0.3)^{0}.(0.7)^{14} = 0.0068$

$P(X = 1) = C_{14,1}.(0.3)^{1}.(0.7)^{13} = 0.0407$

$P(X = 2) = C_{14,2}.(0.3)^{2}.(0.7)^{12} = 0.1134$

$P(X = 3) = C_{14,3}.(0.3)^{3}.(0.7)^{11} = 0.1943$

$P(X = 4) = C_{14,4}.(0.3)^{4}.(0.7)^{10} = 0.2290$

$P(X = 5) = C_{14,5}.(0.3)^{5}.(0.7)^{9} = 0.1963$

$P(X = 6) = C_{14,6}.(0.3)^{6}.(0.7)^{8} = 0.1262$

$P(X < 7) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) = 0.0068 + 0.0407 + 0.1134 + 0.1943 + 0.2290 + 0.1963 + 0.1262 = 0.9067$

90.67% probability that John finds less than 7 golden sheets of paper

7 0

3 years ago

A student was asked to use the formula for the perimeter of a rectangle, p = 2l 2w, to solve for l. the student came up with an

timofeeve [1]

<span>A student was asked to use the formula for the perimeter of a rectangle, p = 2l 2w, to solve for l. the student came up with an answer, p -2w=2l. what</span>

6 0

3 years ago

Amrita thinks of a number. She doubles it, adds 9 , divides her answer by 3 and finally subtracts 1 . She obtains the same numbe

maks197457 [2]

Answer:

Amrita's number is 6

Step-by-step explanation:

x Double the unknown number

2x Add 9

2x + 9 Divide the answer by 3

2/3x + 3 Subtract 1

2/3x + 2 Set this equal to the original number

2/3x + 2 = x

-2/3x -2/3x Subtract 2/3x from both sides

2 = 1/3x Multiply both sides by 3

6 = x

5 0

3 years ago

Simplify the expression-2.3 + (-5.7). On the test when maureen simplified the expression she got -3.4. What mistake did maureen

Hoochie [10]

Answer:

see explanation

Step-by-step explanation:

- 2.3 + (- 5.7)

reminder that + (- ) = -

To obtain - 3.4 it is likely that she added 2.3 to - 5.7

The solution is

- 2.3 - 5.7 = - 8

4 0

3 years ago

Read 2 more answers

Many universities and colleges have instituted supplemental instruction (SI) programs, in which a student facilitator meets regu

aev [14]

Answer:

Step-by-step explanation:

Let many universities and colleges have conducted supplemental instruction(SI) programs. In that a student facilitator he meets the students group regularly who are enrolled in the course to promote discussion of course material and enhance subject mastery.

Here the students in a large statistics group are classified into two groups:

1). Control group: This group will not participate in SI and

2). Treatment group: This group will participate in SI.

a)Suppose they are samples from an existing population, Then it would be the population of students who are taking the course in question and who had supplemental instruction. And this would be same as the sample. Here we can guess that this is a conceptual population - The students who might take the class and get SI.

b)Some students might be more motivated, and they might spend the extra time in the SI sessions and do better. Here they have done better anyway because of their motivation. There is other possibility that some students have weak background and know it and take the exam, But still do not do as well as the others. Here we cannot separate out the effect of the SI from a lot of possibilities if you allow students to choose.

The random assignment guarantees ‘Unbiased’ results - good students and bad are just as likely to get the SI or control.

c)There wouldn't be any basis for comparison otherwise.

5 0

3 years ago

93. Let f(x) = 3x - 1 and g(x) = x + 5. Evaluate the following limits. (a) lim x→3 f(g(x)] (b) lim x→3 g[f(x)]​

93. Let f(x) = 3x - 1 and g(x) = x + 5. Evaluate the following limits. (a) lim x→3 f(g(x)] (b) lim x→3 g[f(x)]