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3 years ago

Work out the perimeter of a quarter-circle with radius 7cm.

Mathematics

2 answers:

Flura [38]3 years ago

7 0

Answer:

Step-by-step explanation:

p=1÷4pir+2r

1÷4×22÷7×7+2(7)

=11cm+14cm

=25

patriot [66]3 years ago

6 0

Answer:

3.5π cm²

Step-by-step explanation:

let P represent the perimeter of a circle with radius 7

P = 2×π×r

= 2×π×7

= 14π

the perimeter of a quarter-circle with radius 7 = P÷4

= 14π÷4

= 3.5π

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Lelechka [254]

Answer:

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Step-by-step explanation:

if ur supposed to add them then

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3 years ago

m∠A = 42 and m∠C = 57. Find m∠B. A. 81 B. 99 C. 132 D. 147

faltersainse [42]

A.
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3 years ago

Find the area of the triangle.

mars1129 [50]

Answer:

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Step-by-step explanation:

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3 years ago

Chad bought 7 pounds of strawberries for $10.80. What was the cost, c, of 1 pound of strawberries?

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3 years ago

Read 2 more answers

It appears that people who are mildly obese are less active than leaner people. One study looked at the average number of minute

Molodets [167]

Answer:

10.38% probability that the mean number of minutes of daily activity of the 6 mildly obese people exceeds 410 minutes.

99.55% probability that the mean number of minutes of daily activity of the 6 lean people exceeds 410 minutes

Step-by-step explanation:

The Central Limit Theorem estabilishes that, for a random variable X, with mean $\mu$ and standard deviation $\sigma$ , a large sample size can be approximated to a normal distribution with mean $\mu$ and standard deviation $\frac{\sigma}{\sqrt{n}}$ .

Normal probability distribution

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Mildly obese

Normally distributed with mean 375 minutes and standard deviation 68 minutes. So $\mu = 375, \sigma = 68$

What is the probability (±0.0001) that the mean number of minutes of daily activity of the 6 mildly obese people exceeds 410 minutes?

So $n = 6, s = \frac{68}{\sqrt{6}} = 27.76$

This probability is 1 subtracted by the pvalue of Z when X = 410.

$Z = \frac{X - \mu}{s}$

$Z = \frac{410 - 375}{27.76}$

$Z = 1.26$

$Z = 1.26$ has a pvalue of 0.8962.

So there is a 1-0.8962 = 0.1038 = 10.38% probability that the mean number of minutes of daily activity of the 6 mildly obese people exceeds 410 minutes.

Lean

Normally distributed with mean 522 minutes and standard deviation 106 minutes. So $\mu = 522, \sigma = 106$