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Tems11 [23]
3 years ago
15

Calculate the surface area of the figure below.

Mathematics
1 answer:
valentinak56 [21]3 years ago
8 0

Answer:

a. 1,275.49 m²

Step-by-step explanation:

SA = 2πr² + 2πrh

SA = 2π(7)² + 2π(7)(22)

SA = 98π + 308π

SA = 406π

SA ≈ 1,275.49 m²

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Amanda exercised for 10 minutes every day in the first week, 20 minutes in the second week, 30 minutes in the third week, and 40
Dominik [7]

Amanda's method is linear because the number of minutes increased by an equal number every week.

7 0
2 years ago
Identify the like terms 3x,4y,4z,5x<br><br> A:4y,5x<br> B:4y,4z<br> C:3x,5x<br> D:3x,4z
fomenos

Answer:

c

Step-by-step explanation:


5 0
3 years ago
The mean life of a television set is 119 months with a standard deviation of 14 months. If a sample of 74 televisions is randoml
irina [24]

Answer:

50.34% probability that the sample mean would differ from the true mean by less than 1.1 months

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution:

Problems of normally distributed samples are 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.

Central limit theorem:

The Central Limit Theorem estabilishes that, for a random variable X, with mean \mu and standard deviation \sigma, the sample means with size n of at least 30 can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}

In this problem, we have that:

\mu = 119, \sigma = 14, n = 74, s = \frac{14}{\sqrt{74}} = 1.63

If a sample of 74 televisions is randomly selected, what is the probability that the sample mean would differ from the true mean by less than 1.1 months

This is the pvalue of Z when X = 119 + 1.1 = 120.1 subtracted by the pvalue of Z when X = 119 - 1.1 = 117.9. So

X = 120.1

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

By the Central Limit Theorem

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

Z = \frac{120.1 - 119}{1.63}

Z = 0.68

Z = 0.68 has a pvalue of 0.7517

X = 117.9

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

Z = \frac{117.9 - 119}{1.63}

Z = -0.68

Z = -0.68 has a pvalue of 0.2483

0.7517 - 0.2483 = 0.5034

50.34% probability that the sample mean would differ from the true mean by less than 1.1 months

8 0
3 years ago
If you know any of these could yoy hep me out :D
Mekhanik [1.2K]
7. X=5
8. X=6
9. B=-3
10. G=3.4
5 0
3 years ago
5x88x9x7x3x9x10x31x47=
Gre4nikov [31]
1.09047708 E10 i guess
6 0
3 years ago
Read 2 more answers
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