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jeka57 [31]
3 years ago
13

Angelita and Cynthia like to play a video game against each other that is scored by round. The goal is to score as many points a

s possible in each round. Here are summary statistics how many points they each score per round:
Mean Standard deviation
Angelita \mu_A=900μ
A
​
=900mu, start subscript, A, end subscript, equals, 900 \sigma_A=96σ
A
​
=96sigma, start subscript, A, end subscript, equals, 96
Cynthia \mu_C=800μ
C
​
=800mu, start subscript, C, end subscript, equals, 800 \sigma_C=72σ
C
​
=72sigma, start subscript, C, end subscript, equals, 72
Both distributions are approximately normal. Suppose we choose a round at random and calculate the difference between their scores. We can assume that their scores each round are independent.
Find the probability that Angelita's score is higher than Cynthia's.
You may round your answer to two decimal places.
P\left(\text{Angelita higher}\right)\approxP(Angelita higher)≈
Mathematics
1 answer:
ladessa [460]3 years ago
8 0

Answer:

0.80

Step-by-step explanation:

Khan academy

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The dot plots show rainfall totals for several spring storms in highland areas and lowland areas. What is the median rainfall fo
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Answer:

Median highland = 16 mm

Median lowland = 16 mm

Step-by-step explanation:

For the highland area :

The sample size, n = 44

Obtain the median value using :

1/2 * (n + 1)th term

1/2 * (44 +1) th term

1/2 * 45 = 22.5th term.

(22nd + 23rd) term / 2

(16 + 16) / 2 = 16 mm

For the lowland area :

1/2 * (n + 1)th term

1/2 * (44 +1) th term

1/2 * 45 = 22.5th term.

(22nd + 23rd) term / 2

(12 + 12) / 2 = 12 mm

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3 years ago
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Which pairs of polygons are similar? Select each correct answer. Two trapezoids. The trapezoid on the left has horizontal bases
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Answer:

*thinking*

Step-by-step explanation:

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8 0
3 years ago
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List the perfect cubes between 1000 and 3000 that are even numbers
ad-work [718]
 12^3 = 1728; 14^3 = 2744;
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The management of a relatively new social networking website named BooglePlus is conducting a pilot study comparing use of its o
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Answer:

0.0623 ± ( 2.056 )( 0.0224 ) can be used to compute a 95% confidence interval for the slope of the population regression line of y on x

Step-by-step explanation:

Given the data in the question;

sample size n = 28

slope of the least squares regression line of y on x or sample estimate = 0.0623

standard error = 0.0224

95% confidence interval

level of significance ∝ = 1 - 95% = 1 - 0.95 = 0.05

degree of freedom df = n - 2 = 28 - 2 = 26

∴ the equation will be;

⇒ sample estimate ± ( t-test) ( standard error )

⇒ sample estimate ± ( t_{\alpha /2, df) ( standard error )

⇒ sample estimate ± ( t_{0.05 /2, 26) ( standard error )

⇒ sample estimate ± ( t_{0.025, 26) ( standard error )

{ from t table; ( t_{0.025, 26) = 2.055529 = 2.056

so we substitute

⇒ 0.0623 ± ( 2.056 )( 0.0224 )

Therefore, 0.0623 ± ( 2.056 )( 0.0224 ) can be used to compute a 95% confidence interval for the slope of the population regression line of y on x

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3 years ago
103,727,495 in a word form
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103,727,495 in word form is: one hundred three million, seven hundred twenty-seven thousand, four hundred ninety-five.
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