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3 years ago

14

You have sample data that leads you to believe that the average high temperature in Oklahoma City is 92°F.

2 answers:

Sliva [168]3 years ago

8 0

Answer:

B - u>90 on edge

valentinak56 [21]3 years ago

4 0

Answer:

B. µ > 90

Step-by-step explanation:

I can confirm that the answer to your question is B.

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If sin theta = -1/2 and lies in the third quadrant, what is the cos?

never [62]

Sin angle that equals the 1/2 value is the 30 degree on the 1st Quadrant.

The angle of 30 degree on the 3rd Quadrant is equivalent to (formula and circle in the 1st photo) 180 + 30 = 210°.

So theta = 210° (3rd Quadrant) = 30° (1st Quadrant).

$cos \: 30\degree = \sqrt{3} \div 2$

Now just transfer the cos 30° to the 3rd Quadrant and the signal will be negative (as shown in the 2nd photo attached).

$cos \: 210 \degree = - \sqrt{3} \div 2$

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3 years ago

Jacob and his team plays third in this years Ultimate frisbee tournament the o official frisbee be used in the tournament had a

tankabanditka [31]

The circumference is 31.4

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3 years ago

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These cones are similar. Find the volume

Serga [27]

Answer:

the answer is 44cm3

Step-by-step explanation:

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1 year ago

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Look at the table of values below.

Eduardwww [97]

Answer:

y = 2x - 1

Step-by-step explanation:

1(2) - 1 = 1

2(2) - 1 = 3

3(2) - 1 = 5

4(2) - 1 = 7

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2 years ago

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132 days to 125.4 days percent of change

Evgen [1.6K]

Answer:

132 days to 125.4 days percent of change

Step-by-step explanation:

p= N-O/O x 100% where p is the percent change, is the New value and O is

the Old value. We are given the Old value (132) and the New value (125.4) which we can substitute into the formula to calculate the percent change:

p= 125.4-132/132 x 100%

p= -6.6/132 x 100%

p= (-660/132%)

p= -5%

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7 months ago

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