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
spin [16.1K]
3 years ago
10

HELP ME PLEASE QUICKLY

Mathematics
2 answers:
GenaCL600 [577]3 years ago
8 0
1+5(divided by)5
Since the problem asks to use PEMDAS, you divide 5 by 5 FIRST:
1+1
Then add:
2

Your answer is 2.
katrin2010 [14]3 years ago
4 0
Step one: plug in the # so 1 + 5/5
using pemdas you would dived the 5's first so then it would be 1+1 = 2
i hoped that helped ! 

You might be interested in
Plz help number 4 plz rn
ruslelena [56]
The answer for this qeustion is C.
7 0
3 years ago
Read 2 more answers
For me and Thomas uhh
atroni [7]

Answer:

what does that question mean?

5 0
3 years ago
Please creeper keeps answering but link doesnt work plaese help
DedPeter [7]

Answer:

The answer is A: y^2/2x

Step-by-step explanation:

Hope this helps please mark brainliest  :)

6 0
3 years ago
There are two college entrance exams that are often taken by students, Exam A and Exam B. The composite score on Exam A is appro
elena55 [62]

Answer:

B.The score on Exam A is better, because the percentile for the Exam A score is higher.

Step-by-step explanation:

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

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

The Z-score 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. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

Two exams. The exam that you did score better is the one in which you had a higher zscore.

The composite score on Exam A is approximately normally distributed with mean 20.1 and standard deviation 5.1.

This means that \mu = 20.1, \sigma = 5.1.

You scored 24 on Exam A. So

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

Z = \frac{24 - 20.1}{5.1}

Z = 0.76

The composite score on Exam B is approximately normally distributed with mean 1031 and standard deviation 215.

This means that \mu = 1031, \sigma = 215.

You scored 1167 on Exam B, s:

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

Z = \frac{1167 - 1031}{215}

Z = 0.632

You had a better Z-score on exam A, so you did better on that exam.

The correct answer is:

B.The score on Exam A is better, because the percentile for the Exam A score is higher.

3 0
4 years ago
Which of the following activities are examples of data gathering? Check all that apply. A. Surveying students about their favori
katrin [286]
A. Surveying the students about their favorite type of math problem.
and
C. Sampling  water for contaminants
6 0
3 years ago
Read 2 more answers
Other questions:
  • Double the product of u and 7
    6·1 answer
  • Find the value of 3.4 squared
    14·2 answers
  • A cylinder has a height of 16 cm and a radius of 5 cm. A cone has a height of 12 cm and a radius of 4 cm. If the cone is placed
    14·1 answer
  • The quotient of a number and 3 less than twice the number
    10·1 answer
  • Suppose M is the midpoint of segment CD, CM = 5x-12 and MD = 2x + 9. What is x?​
    7·1 answer
  • write a system of equations for the problem, and then solve the system. if a plane can travel 340 miles per hour with the wind o
    9·1 answer
  • Given: ∠1 and ∠2 are complementary angles.<br>Prove: ∠3 and ∠4 are complementary angles.
    8·2 answers
  • Help help help help help
    7·1 answer
  • In the ordered pair (2, 3), what is the x-coordinate? y-coordinate?
    15·2 answers
  • Find the value of x
    7·2 answers
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!