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gavmur [86]
2 years ago
14

A golfer recorded the following scores for each of four rounds of golf: 86, 81, 87, 82. The mean of the scores is 84. What is th

e sum of the squared deviations of the scores from the mean?
Mathematics
1 answer:
Bezzdna [24]2 years ago
3 0

Answer:

26

Step-by-step explanation:

Subtract the mean (84) from each of the four scores, obtaining

2, -3, 3, - 2

These are the "deviations."

Next, square each of these four deviations, obtaining

(2)^2 = 4, and:

(-3)^2 = 9, and:

(3)^2 = 9, and:

(-2)^2 = 4

Then the sum of the squared deviations of the scores from the mean is 26.

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Answer:

a) x-4+9

b) x-2

For part b, I am not 100% sure about my answer, but I am sure about part a.

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Simplify x + 2/ x^2 - 6x - 16 divided by 1/9x
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Alexandra's bank statement shows a closing balance of $130.68. There are no outstanding checks or deposits. Her checkbook shows
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5 0
2 years ago
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For questions 2-5, the number of pieces in a regular bag of Skittles is approximately normally distributed with a mean of 38.4 a
aleksley [76]

Answer:

a. -1.60377

b. 0.25451

c. 0.344

d. Option b) 78th

Step-by-step explanation:

The number of pieces in a regular bag of Skittles is approximately normally distributed with a mean of 38.4 and a standard deviation of 2.12.

a)What is the z-score value of a randomly selected bag of Skittles that has 35 Skittles? a) 1.62 b) -1.62 c) 3.40 d) -3.40 e)1.303.

The formula for calculating a z-score is is z = (x-μ)/σ,

where x is the raw score

μ is the population mean

σ is the population standard deviation.

z = 35 - 38.4/2.12

= -1.60377

Option b) -1.62 is correct

b) What is the probability that a randomly selected bag of Skittles has at least 37 Skittles? a) .152 b) .247 c) .253 d).747e).7534. .

z = (x-μ)/σ

Mean of 38.4 and a standard deviation of 2.12.

z = (37 - 38.4)/2.12

= -0.66038

P-value from Z-Table:

P(x<37) = 0.25451

The probability that a randomly selected bag of Skittles has at least 37 Skittles is 0.25451

Option c) .253 is.correct

c) What is the probability that a randomly selected bag of Skittles has between 39 and 42 Skittles? a) .112 b) .232 c) .344 d).457 e).6125.

z = (x-μ)/σ

Mean of 38.4 and a standard deviation of 2.12.

For 39 Skittles

z = (39 - 38.4)/2.12

= 0.28302

Probability value from Z-Table:

P(x = 39) = 0.61142

For 42 Skittles

z = (42 - 38.4)/2.12

= 1.69811

Probability value from Z-Table:

P(x = 42) = 0.95526

The probability that a randomly selected bag of Skittles has between 39 and 42 Skittles is:

P(x = 42) - P(x = 39

0.95526 - 0.61142

0.34384

= 0.344

Option c is.correct

d) What is the percentile rank of a randomly selected bag of Skittles that has 40 Skittles in it? a)82nd b) 78th c) 75th d)25th e)22nd

z = (x-μ)/σ

Mean of 38.4 and a standard deviation of 2.12.

z = (40 - 38.4)/2.12

= 0.75472

P-value from Z-Table:

P(x = 40) = 0.77479

Converting to percentage = 0.77479× 100

= 77. 479%

≈ 77.5

Percentile rank = 78th

7 0
2 years ago
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Answer:

Leon is correct. (Option 1)

Step-by-step explanation:

Given that Leon verified that the side lengths 21, 28, 35 form a Pythagorean triple using this procedure.

Step 1: Find the greatest common factor of the given lengths: 7

Step 2: Divide the given lengths by the greatest common factor: 3, 4, 5

Step 3: Verify that the lengths found in step 2 form a Pythagorean triple.

we have to explain whether or not Leon is correct.

As, 3,4,5 forms a Pythagorean triplet i.e satisfies the Pythagoras theorem

Hypotenuse^2=Base^2+Perpendicular^2

⇒ 5^2=3^2+4^2

Let a, b, c forms a Pythagorean triplet

a^2+b^2=c^2

Multiplied by 4 on both sides

⇒ 4a^2+4b^2=4c^2

⇒ {2a}^2+{2b}^2={2c}^2

Hence, we say 4a, 4b and 4c also forms a Pythagorean triplet.

∴ multiplying every length of a Pythagorean triple by the same whole number results in a Pythagorean triple.

Hence, Leon is correct.

4 0
2 years ago
Read 2 more answers
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