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Sphinxa [80]
4 years ago
9

Find the factored form -13m-m^2=0

Mathematics
1 answer:
mash [69]4 years ago
7 0
M=0, -13


hope that helps! :)
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Erica plotted the three towns closest to her house on a graph with town A
marta [7]
First, determine the distances from the towns.
  Town A to Town B: Opposite angle C
                      d = sqrt ((12 - 7)^2 + (9 - 9)^2)
                      d = 5
  Town B to Town C: Opposite Angle A
                      d = sqrt ((7 - 1)^2 + (9 - 1)^2)
                       d = 10
  Town A to Town C: Opposite Angle B
                       d = sqrt (12 - 1)^2 + (9 - 1)^2)
                        d = 13.6
The angle with largest measure is the one opposite the greatest distance. Thus, the arrangement would be:    C, A, B.
8 0
3 years ago
The box plots below show the ages of college students in different math courses.
Slav-nsk [51]

Answer:

The correct option for this is C.) The mean and median age are more likely to be the same for the students in Math 1.

Step-by-step explanation:

i) The median age of the students is Math 1 is less than the median age of the students in Math 2.

This statement is NOT TRUE as the median age of students in Math 1 = median age of students in Math 2 = 19

ii) The mean and median age are most likely the same for both sets of data.

   This statement is NOT TRUE. The mean age of students in Math 2 should be greater than mean age of students in Math 1.

iii) The mean and median age are more likely to be the same for the students in Math 1.

  This statement is TRUE.

8 0
3 years ago
Read 2 more answers
H - 4/ j= k for J please help
dimulka [17.4K]

h=k+4/j dude just go this thing called cymath


6 0
3 years ago
What is 6+5342 equal
ASHA 777 [7]

5.348

this to weird, where is the trap?

5 0
3 years ago
Read 2 more answers
Refer to the Teaching Psychology (May 1998) study of how external clues influence performance. Two different forms of a midterm
kozerog [31]

Answer:

P(X

P(Y

So for this case we have a higher probability for the red exam so we can conclude that the student is more likely to score below 20% on the difficult questions in the red one.

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

Solution to the problem

Let X the random variable that represent the scores for the blue exam, and for this case we know the distribution for X is given by:

X \sim N(53,15)  

Where \mu=53 and \sigma=15

We are interested in the probability that P(X<20%)

And the best way to solve this problem is using the normal standard distribution and the z score given by:

z=\frac{x-\mu}{\sigma}

And if we replace we got:

P(X

Let Y the random variable that represent the scores for the red exam, and for this case we know the distribution for X is given by:

Y \sim N(39,12)  

Where \mu=39 and \sigma=12

We are interested in the probability that P(Y<20%)

And the best way to solve this problem is using the normal standard distribution and the z score given by:

z=\frac{y-\mu}{\sigma}

And if we replace we got:

P(Y

So for this case we have a higher probability for the red exam so we can conclude that the student is more likely to score below 20% on the difficult questions in the red one.

7 0
4 years ago
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