Which could not be a set of intervals for a frequency table or a histogram? Question 1 options:

0–4, 5–9, 10–14, 15–19, 20–24

1–5, 6–10, 11–15, 16–20, 21–25

1–10, 11–20, 21–30, 31–40, 41–50

0–5, 5–9, 10–15, 15–20, 20–25

second option

step-by-step explanation:

the measures of a triangle equal 180. use that to find x.

2x+3x+4x=180

9x=180

9x/9=180/9

x=20

now you know x, you know angle a is 40°, angle b is 60°, angle c is 80°.

because angle a is 40°, the first option doesn't work.

because angle b is 60°, the second option does work.

when you add angle a and b, you get 100°. complementary means it equals 90°, so the third option doesn't work.

when you add angle a and c, you get 120°, which is not 100.

the number of possible outcomes that do not show a 1 on the first spinner and show the number 4 on the second spinner is 2.

spinner 1 spinner 2

1                     1

1                     2

1                     3

1                     4

2                   1

2                   2

2                   3

2                   4   = not 1, and 4

3                   1

3                   2

3                   3

3                   4 = not 1, and 4

only 2 outcomes have passed the given condition.

step-by-step explanation: