1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
nlexa [21]
3 years ago
8

Question 2 (1 point)

Mathematics
1 answer:
timama [110]3 years ago
3 0
42 square cm

Width=6
Length =7
You might be interested in
The distribution of SAT II Math scores is approximately normal with mean 660 and standard deviation 90. The probability that 100
gayaneshka [121]

Using the <em>normal distribution and the central limit theorem</em>, it is found that there is a 0.1335 = 13.35% probability that 100 randomly selected students will have a mean SAT II Math score greater than 670.

<h3>Normal Probability Distribution</h3>

In a normal distribution with mean \mu and standard deviation \sigma, the z-score of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

  • It measures how many standard deviations the measure is from the mean.
  • After finding the z-score, we look at the z-score table and find the p-value associated with this z-score, which is the percentile of X.
  • By the Central Limit Theorem, the sampling distribution of sample means of size n has standard deviation s = \frac{\sigma}{\sqrt{n}}.

In this problem:

  • The mean is of 660, hence \mu = 660.
  • The standard deviation is of 90, hence \sigma = 90.
  • A sample of 100 is taken, hence n = 100, s = \frac{90}{\sqrt{100}} = 9.

The probability that 100 randomly selected students will have a mean SAT II Math score greater than 670 is <u>1 subtracted by the p-value of Z when X = 670</u>, hence:

Z = \frac{X - \mu}{\sigma}

By the Central Limit Theorem

Z = \frac{X - \mu}{s}

Z = \frac{670 - 660}{9}

Z = 1.11

Z = 1.11 has a p-value of 0.8665.

1 - 0.8665 = 0.1335.

0.1335 = 13.35% probability that 100 randomly selected students will have a mean SAT II Math score greater than 670.

To learn more about the <em>normal distribution and the central limit theorem</em>, you can take a look at brainly.com/question/24663213

7 0
2 years ago
A plumber is replacing a pipe in a kitchen. The old pipe
just olya [345]

Answer:

Step-by-step explanation:

Fg

5 0
3 years ago
Solve the system of equations by elimination:
Angelina_Jolie [31]

Answer:

B (1/2,-2)

Step-by-step explanation:

4b + 13c = -24 --(1)

12b - 5c = 16 --(2)

(1)x3

12b + 39c = -72 --(3)

(3)-(2)

12b + 39c - 12b + 5c = -72-16

c = -2

Sub c = -2 into (1)

4b = -24 - 13(-2)

b = 1/2

7 0
3 years ago
The function f(x)=2+e^x has an asymptote at ________.
ElenaW [278]
<h3>Answer: Horizontal asymptote at y = 2</h3>

As x approaches negative infinity, the e^x term slowly approaches 0

This means 2+e^x approaches 2+0 = 2

So that's why we have a horizontal asymptote at y = 2

The graph approaches the line y = 2 but doesn't touch or cross it. Think of it like an electric fence.

The graph is below.

5 0
3 years ago
Tickets of a program at college cost ​$3 for general admission or ​$2 with a student ID. If 181 people paid to see a performance
Makovka662 [10]

Answer:

490

Step-by-step explanation:

3 0
3 years ago
Other questions:
  • What is 45 divided by 6
    8·2 answers
  • PLEASE HELP MMEEEEE BY simplifying 154/308
    10·2 answers
  • The mean length of 258 pages. The longest book has 294 pages. What is the mean length of the other six books
    5·1 answer
  • PLEASE HELP
    8·2 answers
  • How can x + 4 = 3x + 12 be set up as a system of equations?
    11·2 answers
  • Clip 159 grade 5 questions worksheet answers
    10·1 answer
  • A kindergartden class has 9 girls and 11 boys in class . what is the percent of the class is girls
    13·2 answers
  • Which equation shows the
    5·2 answers
  • What the solution to -8+x=3
    10·1 answer
  • Triangle ABC has vertices at (-4,0), (-1,6) and (3,-1). What is the perimeter of Triangle ABC, rounded to the nearest tenth?
    7·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!