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Alona [7]
4 years ago
13

The circumference of a circle is 9.42 cm. What is the area?

Mathematics
1 answer:
VikaD [51]4 years ago
6 0

Answer:7.06in with a 2 squared or exponent

Step-by-step explanation:

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The heights of students at a college are normally distributed with a mean of 175 cm and a standard deviation of 6 cm. One might
frosja888 [35]

Answer:

25

Step-by-step explanation:

The Empirical Rule states that, for a normally distributed random variable:

68% of the measures are within 1 standard deviation of the mean.

95% of the measures are within 2 standard deviation of the mean.

99.7% of the measures are within 3 standard deviations of the mean.

In this problem, we have that:

Mean = 175cm

Standard deviation = 6 cm

Percentage of students below 163 cm

163 = 175 - 2*6

So 163 is two standard deviations below the mean.

By the Empirical rule, 95% of the heights are within 2 standard deviations of the mean. The other 100-95 = 5% are more than 2 standard deviations of the mean. Since the normal distribution is symmetric, 2.5% of them are more than 2 standard deviations below the mean(so below 163cm) and 2.5% are more than two standard deviations above the mean.

2.5% of the students have heights less than 163cm.

Out of 1000

0.025*1000 = 25

25 is the answer

8 0
3 years ago
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∠A and ∠B\angle B∠B are supplementary angles. If m∠A=(8x−27)∘\angle A=(8x-27)^{\circ}∠A=(8x−27)∘ and m∠B=(4x+3)∘\angle B=(4x+3)^
hoa [83]

Answer:< B = 15°

Step-by-step explanation:

If m\angle A=(x+16)^{\circ}∠A=(x+16)​​ and m\angle B=(3x-14)^{\circ}∠B=(3x−14), then find the measure of \angle A∠A.

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3 years ago
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Draw the line of reflection that reflect quadrilateral ABCD onto quadrilateral A'B'C'D
lisov135 [29]

Answer:

-3 across the x

Step-by-step explanation:

negative 3 is in the middle of the two, both between point -1 and -5, and point 0 and -6.

4 0
3 years ago
Please help me idk this
marusya05 [52]

Answer:

The answer would be 30m squared.

Step-by-step explanation:

Multiply side and base and diveide by two.

6 0
3 years ago
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