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nekit [7.7K]
3 years ago
10

In a sale, all prices are reduced by 25%. Work out the

Mathematics
1 answer:
Leno4ka [110]3 years ago
6 0

Answer:

1650

Step-by-step explanation:

2200*.25=1650 sale price

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The distribution of SAT II Math scores is approximately normal with mean 660 and standard deviation 90. The probability that 100
gayaneshka [121]

Using the <em>normal distribution and the central limit theorem</em>, it is found that there is a 0.1335 = 13.35% probability that 100 randomly selected students will have a mean SAT II Math score greater than 670.

<h3>Normal Probability Distribution</h3>

In a normal distribution with mean \mu and standard deviation \sigma, the z-score of a measure X is given by:

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

  • It 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, which is the percentile of X.
  • By the Central Limit Theorem, the sampling distribution of sample means of size n has standard deviation s = \frac{\sigma}{\sqrt{n}}.

In this problem:

  • The mean is of 660, hence \mu = 660.
  • The standard deviation is of 90, hence \sigma = 90.
  • A sample of 100 is taken, hence n = 100, s = \frac{90}{\sqrt{100}} = 9.

The probability that 100 randomly selected students will have a mean SAT II Math score greater than 670 is <u>1 subtracted by the p-value of Z when X = 670</u>, hence:

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

By the Central Limit Theorem

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

Z = \frac{670 - 660}{9}

Z = 1.11

Z = 1.11 has a p-value of 0.8665.

1 - 0.8665 = 0.1335.

0.1335 = 13.35% probability that 100 randomly selected students will have a mean SAT II Math score greater than 670.

To learn more about the <em>normal distribution and the central limit theorem</em>, you can take a look at brainly.com/question/24663213

7 0
2 years ago
The average amount of rent paid for apartments in Campbell is $1,331, which is 10% more than it was last spring. What was the av
baherus [9]

Answer:

1197.9

Step-by-step explanation:

10 percent of 1331 is 133.1.

because its 10 percent more, you need to do 1331-133.1:)

4 0
3 years ago
The owner of a fish market has an assistant who has determined that the weights of catfish are normally distributed, with mean o
prohojiy [21]

Answer:

The percentage of samples of 4 fish will have sample means between 3.0 and 4.0 pounds is 66.87%

Step-by-step explanation:

For a normal random variable with mean Mu = 3.2 and standard deviation sd = 0.8 there is a distribution of the sample mean (MX) for samples of size 4, given by:

Z = (MX - Mu) / sqrt (sd ^ 2 / n) = (MX - 3.2) / sqrt (0.64 / 4) = (MX - 3.2) / 0.4

For a sample mean of 3.0, Z = (3 - 3.2) / 0.4 = -0.5

For a sample mean of 3.0, Z = (4 - 3.2) / 0.4 = 2.0

P (3.2 <MX <4) = P (-0.5 < Z <2.0) = 0.6687.

The percentage of samples of 4 fish will have sample means between 3.0 and 4.0 pounds is 66.87%

6 0
3 years ago
Please show work and help!!
grandymaker [24]
A.25
Work:
-5x-5=25
Hope his helps.
4 0
3 years ago
Read 2 more answers
Round 5.659 to the nearest hundredth
Elan Coil [88]

Answer:

Your answer will be 5.66

7 0
3 years ago
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