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iris [78.8K]
2 years ago
13

Jenna and Luke each earned 80 points while playing a game. Luke received 8 additional points for completing the

Mathematics
1 answer:
serg [7]2 years ago
4 0

Answer:

80-2= n

Step-by-step explanation:

jenna has 2 fewer points thefore 80-2= n, n is how much she got ( she got 78 points)

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Answer:
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then you put it in an equation
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If a pair of shoes has an original price of 80 and they are on sale for 48, what is the percent of discount?
defon
60% is the correct answer.
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For f(x) = 3x + 1 and g(x) = x2 - 6, find (f+ g)(x).
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B

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4 0
2 years ago
For a fair coin, suppose you toss the coin 100 times.
Natasha_Volkova [10]

Using the normal distribution, it is found that there is a 0.0005 = 0.05% probability of getting more than 66 heads.

<h3>Normal Probability Distribution</h3>

The z-score of a measure X of a normally distributed variable with mean \mu and standard deviation \sigma is given by:

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

  • The z-score measures how many standard deviations the measure is above or below the mean.
  • Looking at the z-score table, the p-value associated with this z-score is found, which is the percentile of X.
  • The binomial distribution is the probability of x successes on n trials, with p probability of a success on each trial. It can be approximated to the normal distribution with \mu = np, \sigma = \sqrt{np(1-p)}.

For the binomial distribution, the parameters are given as follows:

n = 100, p = 0.5.

Hence the mean and the standard deviation of the approximation are given as follows:

  • \mu = np = 100(0.5) = 50.
  • \sigma = \sqrt{np(1-p)} = \sqrt{100(0.5)(0.5)} = 5

Using continuity correction, the probability of getting more than 66 heads is P(X > 66 + 0.5) = P(X > 66.5), which is <u>one subtracted by the p-value of Z when X = 66.5</u>.

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

Z = \frac{66.5 - 50}{5}

Z = 3.3

Z = 3.3 has a p-value of 0.9995.

1 - 0.9995 = 0.0005.

0.0005 = 0.05%

More can be learned about the normal distribution at brainly.com/question/4079902

#SPJ1

6 0
1 year ago
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