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asambeis [7]
3 years ago
8

Taryn collected data for the rain collected over a number of days in various cities. Which of the data sets represents a scatter

plot with a line of best fit that BEST displays a relationship where, as the number of days increases, the amount of rain collected remains constant?
A.
(5, 1), (2, 7), (3, 6), (4, 4), (1, 8), (6, 0)

B.
(1, 6), (5, 6), (3, 5), (4, 5), (6, 5)

C.
(3, 1), (5, 5), (1, 6), (6, 2), (2, 7)

D.
(2, 3), (6, 8), (5, 8), (1, 1), (3, 5), (4, 6)
Mathematics
1 answer:
garri49 [273]3 years ago
5 0

Answer:

Step-by-step explanation:

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Effectus [21]

Answer:

#9 slope is -2, y int is 3

#10 slope is 3/4, y int is 0

8 0
3 years ago
Giving Brainliest!! 2. Suppose that a frequency histogram and a cumulative frequency histogram are constructed from the same set
Ostrovityanka [42]
<span>The answer would be:
B. The frequency is LESS THAN OR EQUAL TO the cumulative frequency.
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3 0
4 years ago
Read 2 more answers
Help me again, Last question!
storchak [24]

Answer:

C

Step-by-step explanation:

Area of a parallelogram = base times height.

The base is 14 and the height is 3, you just have to rotate the figure.

A = 14 * 3 = 42 square feet

3 0
2 years ago
How many minutes he does arm routine and abdominal routine
AlekseyPX
Let x be the amount of time it takes to perform an arm routine and y be the amount of time it takes to perform an abdominal routine.  We see:

4x+y=70
4x+2y=100

Subtracting the second equation from the first gives y=30.  Substituting gives 4x+30=70, so 4x=40 and x=10.

Thus, an arm routine takes ten minutes and an abdominal routine takes thirty minutes.
5 0
4 years ago
Prove the following statement.
gayaneshka [121]

Answer:

You can prove this statement as follows:

Step-by-step explanation:

An odd integer is a number of the form 2k+1 where k\in \mathbb{Z}. Consider the following cases.

Case 1. If k is even we have: (2k+1)^{2}=(2(2s)+1)^{2}=(4s+1)^{2}=16s^2+8s+1=8(2s^2+s)+1.

If we denote by m=2s^2+2 we have that (2k+1)^{2}=8m+1.

Case 2. if k is odd we have: (2k+1)^{2}=(2(2s+1)+1)^{2}=(4s+3)^{2}=16s^2+24s+9=16s^{2}+24s+8+1=8(2s^{2}+3s+1)+1.

If we denote by m=2s^{2}+24s+1 we have that (2k+1)^{2}=8m+1

This result says that the remainder when we divide the square of any odd integer by 8 is 1.

6 0
3 years ago
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