Help please an equation where the slope is 5 and y-intercept is 12.

Mathematics

1 answer:

Lilit [14]3 years ago

7 0

Answer: y = 5x + 12

Step-by-step explanation:

y = mx + b, where m is the slope, and b is the y- intercept, then we plug in:

y = 5x + 12

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If the level of significance of a hypothesis test is raised from 0.005 to 0.2, the probability of a type ii error will:_________

LiRa [457]

The probability of type II error will decrease if the level of significance of a hypothesis test is raised from 0.005 to 0.2.

<h3 /><h3>What is a type II error?</h3>

A type II error occurs when a false null hypothesis is not rejected or a true alternative hypothesis is mistakenly rejected.

It is denoted by 'β'. The power of the hypothesis is given by '1 - β'.

<h3>How the type II error is related to the significance level?</h3>

The relation between type II error and the significance level(α):

The higher values of significance level make it easier to reject the null hypothesis. So, the probability of type II error decreases.
The lower values of significance level make it fail to reject a false null hypothesis. So, the probability of type II error increases.
Thus, if the significance level increases, the type II error decreases and vice-versa.

From this, it is known that when the significance level of the given hypothesis test is raised from 0.005 to 0.2, the probability of type II error will decrease.

Learn more about type II error of a hypothesis test here:

brainly.com/question/15221256

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7 0

1 year ago

Students in a zoology class took a final exam. They took equivalent forms of the exam at monthly intervals thereafter. After t

juin [17]

Answer:

a. 73; b. 48.9; c. 2; d. 33.8; e. 73

Step-by-step explanation:

Assume the function was

S(t)= 73 - 15 ln(t + 1), t ≥ 0

a. Average score at t = 0

S(0) = 73 - 15 ln(0 + 1) = 73 - 15 ln(1) = 73 - 15(0) =73 - 0 = 73

b. Average score at t = 4

S(4) = 73 - 15 ln(4 + 1) = 73 - 15 ln(5) = 73 - 15(1.61) =73 - 24.14 = 48.9

c. Average score at t =24

S(24) = 73 - 15 ln(24 + 1) = 73 - 15 ln(25) = 73 - 15(3.22) =73 - 48.28 = 24.7

d. Percent of answers retained

At t = 0. the students retained 73 % of the answers.

At t = 24, they retained 24.7 % of the answers.

$\text{Percent retention} = \dfrac{\text{24.7}}{\text{73}} \times 100 \, \% = \text{33.8 \%}\\\\\text{The students retained $\large \boxed{\mathbf{33.8 \, \%}}$ of their original knowledge after two years.}$