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castortr0y [4]
2 years ago
15

There are 120 seventh graders at Danielle's school. Each day, the students have the option of eating outside or in the lunch roo

m. Danielle decided to keep track of how many students eat outside each day. After recording her results for the whole school year, she has displayed the data for each month in a box-and-whisker plot. Below are her plots for the months of September and April.
-Fill in the blank-
The____number of students that ate outside in
mode, mean, median
April is approximately___
25, 10, 15 ,20
___Than the
greater, less
____number of students that ate outside in September.
median, mode, mean
There is more variability in the plot representing_____.
April, September
There is ____overlap in the two data sets.
much, some, no

Mathematics
2 answers:
Oksanka [162]2 years ago
8 0

Answer:

mean, 15, less, median, april, much

Step-by-step explanation:

Katena32 [7]2 years ago
4 0

Answer:

mean

15

less

median

april

much

Step-by-step explanation:

give the other brainly :)

You might be interested in
Choose all that correctly solve for x.
Tamiku [17]

Answer:

Options: A, B, and C correctly solve for x.

Step-by-step explanation:

A).

\frac{5}{2} x = \frac{15}{2}

Multiplying both sides by 2 gives;

5x = 15

x = 15 ÷ 5 = 3

∴ This option correctly solve for x.

B).

x + \frac{7}{2} = 7

x = 7 - \frac{7}{2} = \frac{14}{2} -\frac{7}{2} = \frac{7}{2}

∴ This option correctly solve for x.

C).

x + 3 = \frac{21}{5}

x = \frac{21}{5} - \frac{15}{5}  = \frac{6}{5}

But the option give x as 5/6 hence this option does not correctly solve for x.

D).

5x = 11/2

x = 11/2 ÷ 5 = 11/2 × 1/5 = 11/10

But the option gives x as 10/11 so it does not correctly solve for x.

8 0
3 years ago
The area of a rectangular portrait is 20 square feet. It is 4 feet wide. How high is it?
vladimir2022 [97]

Answer:

5 feet high

Step-by-step explanation:

A= LxW

Yx4= 20

  /4    /4

Y= 5ft

It is 5ft high

7 0
2 years ago
Find a 75% confidence interval for the population average weight μ of all adult mountain lions in the specified region. (Round y
kupik [55]

Answer:

75% confidence interval is 91.8±16.66. That is between 75.1 and 108.5 pounds.

Step-by-step explanation:

The question is missing. It is as follows:

Adult wild mountain lions (18 months or older) captured and released for the first time in the San Andres Mountains had the following weights (pounds): 69  104  125  129  60  64

Assume that the population of x values has an approximately normal distribution.

Find a 75% confidence interval for the population average weight μ of all adult mountain lions in the specified region. (Round your answers to one decimal place.)

75% Confidence Interval can be calculated using M±ME where

  • M is the sample mean weight of the wild mountain lions (\frac{69 +104 +125 +129+60 +64}{6} =91.8)
  • ME is the margin of error of the mean

And margin of error (ME) of the mean can be calculated using the formula

ME=\frac{t*s}{\sqrt{N} } where

  • t is the corresponding statistic in the 75% confidence level and 5 degrees of freedom (1.30)
  • s is the standard deviation of the sample(31.4)
  • N is the sample size (6)

Thus, ME=\frac{1.30*31.4}{\sqrt{6} } ≈16.66

Then 75% confidence interval is 91.8±16.66. That is between 75.1 and 108.5

3 0
3 years ago
Someone help me on this one
Murljashka [212]
I’m pretty sure it would be 1/5, because scaling is changing the y values of the function. If the y value of the original is 10, you need to multiply/scale the new one by 1/5 to get the y value of 2
6 0
3 years ago
Read 2 more answers
4 _ (3 _ 2) _ 6 _ 1 = 14 Please help ill give you brainly!
lubasha [3.4K]
4x(3+2)-6x1= 14 is the answer
6 0
2 years ago
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