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Nina [5.8K]
3 years ago
11

Minneapolis, Minnesota and Portland, Oregon have very similar latitudes. (Latitude measures how far north or south a place is, w

hich in turn affects the temperature.)
The histograms below show the daily high temperatures (in degrees Celsius) that were recorded during July 201320132013 in each city.
Which pieces of information can be gathered from these histograms?
Choose all answers that apply:
Choose all answers that apply:

(Choice A)
A
It was warmer in Minneapolis every day.

(Choice B)
B
Minneapolis had the coldest daily high temperature of these two cities.

(Choice C)
C
None of the above
Mathematics
2 answers:
babymother [125]3 years ago
4 0
A it was warmer in Minneapolis every day i think
andreev551 [17]3 years ago
3 0

Answer:

A and C

Step-by-step explanation:

-On average Minneapolis was colder during January

-The daily high temperatures vary more noticeably in Minneapolis

Are BOTH correct.

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Pls answer 10 and 12 show your work
charle [14.2K]
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8 0
4 years ago
Bob and Mark talk about their families. Bob says he has 3 kids, the product of their ages is 72. He gives another clue: the sum
Aloiza [94]
All there ages is 24 all together because if you divide 72/3 you get 24 then multiply it by 3 you will get 72
5 0
3 years ago
Read 2 more answers
Help please... I need help
Contact [7]

Step-by-step explanation:

3x = 12 - 18

3x = (- 6)

X = (-6)/3

x = (-3)

6 0
3 years ago
A parallelogram has a base of 9 km and a height of 5 km.
Luba_88 [7]

Answer:

Area= b*h

  1. 9*5=45km^2

Step-by-step explanation:

6 0
3 years ago
Suppose the number of residents within five miles of each of your stores is asymmetrically distributed with a mean of 25 thousan
ohaa [14]

Answer:

P(\bar X\:>27.8\:thousand)=3.07\%

Step-by-step explanation:

Since the number of residents within five miles of each of your stores is asymmetrically distributed, the distribution of the sample means will be approximately normal with a mean of 25 thousand.

The standard deviation of the sample means is:

\sigma_X=\frac{\sigma}{\sqrt{n} }

\implies \sigma_X=\frac{10.6}{\sqrt{50} }=1.4991

The z value is z=\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n}} }

We plug in the values to get:

z=\frac{27.8-25}{1.4991}=\frac{2.8}{1.4991}=1.87

The area to the right of 1.87 is 1-0.96926=0.03074.

The probability that the average number of residents within five miles of each store in a sample of 50 stores will be more than 27.8 thousand is 3.07%

See attachment.

6 0
4 years ago
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