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Gnesinka [82]
3 years ago
13

Scores turned in by an amateur golfer at a golf course during 2017 and 2018 are as follows.

Mathematics
1 answer:
SSSSS [86.1K]3 years ago
4 0

Answer:

For 2017 Season:

Mean = 75

Standard deviation = 2.07

For 2018 Season:

Mean = 75

Standard deviation = 5.26

Step-by-step explanation:

The following formulas will be used in these calculations:

Mean = (sum of the values) / n

Variance = ((Σ(x - mean)^2) / (n - 1)

Standard deviation = Variance^0.5

Where;

n = number of values = 8

x = each value

For 2017 Season

Mean = (73 + 77 + 78 + 76 + 74 + 72 + 74 + 76) / 8 = 600 / 8 = 75

Variance = ((73-75)^2 + (77-75)^2 + (78-75)^2 + (76-75)^2 + (74-75)^2 + (72-75)^2 + (74-75)^2 + (76-75)^2) / (8-1) = 30 / 7 = 4.29

Standard deviation = Variance^0.50 = 4.29^0.5 = 2.07

For 2018 Season

Mean = (70 + 69 + 74 + 76 + 84 + 79 + 70 + 78) / 8 = 75

Variance = ((70-75)^2 + (69-75)^2 + (74-75)^2 + (76-75)^2 + (84-75)^2 + (79-75)^2 + (70-75)^2 + (78-75)^2) / (8-1) = 194 / 7 = 27.71

Standard deviation = Variance^0.50 = 27.71^0.5 = 5.26

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kupik [55]
A Nonogon has a whopping 9 sides

hope this helps! :D
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3 years ago
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A coffee machine dispenses normally distributed amounts of coffee with a mean of 12 ounces and a standard deviation of 0.2 ounce
NemiM [27]

Answer:

0.9332 = 93.32% probability that the mean of the sample will be less than 12.1 ounces.

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 normal distributions can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the z-score 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 p-value, 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 normally distributed random variable X, with mean \mu and standard deviation \sigma, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

Mean of 12 ounces and a standard deviation of 0.2 ounces.

This means that \mu = 12, \sigma = 0.2

Sample of 9:

This means that n = 9, s = \frac{0.2}{\sqrt{9}}

Find the probability that the mean of the sample will be less than 12.1 ounces.

This is the pvalue of Z when X = 12.1. So

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

By the Central Limit Theorem

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

Z = \frac{12.1 - 12}{\frac{0.2}{\sqrt{9}}}

Z = 1.5

Z = 1.5 has a pvalue of 0.9332

0.9332 = 93.32% probability that the mean of the sample will be less than 12.1 ounces.

6 0
3 years ago
Consider solving the equation for x.
Anna [14]
So let's simplify the equation first.

9x + 12 - 5 = ax + b
9x + 7 = ax + b
9x - ax - b = 7

Alright now plug in the numbers for each option.
I don't feel like doing all of them but I did A (one solution) and B (no solution because the two 9x's cancel out).
6 0
4 years ago
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Will give 5 stars (please answer)
enyata [817]
5 mile 6 mile 8 miles
5 0
3 years ago
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Answer plz...solve this sum n help me​
tamaranim1 [39]

Answer:

28.57 m²

Step-by-step explanation:

Use rational terms.

\frac{\frac{7}{9} }{200} =\frac{\frac{1}{9} }{x}

Cross multiply:

7/9 · x = 1/9 · 200

7/9x = 200/9

x = 200/9 ÷ 7/9

x = 200/9 × 9/7

x = 200 · 9 / 9 · 7

x = 200 / 7

x = 28.57 m

3 0
4 years ago
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