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aleksklad [387]
2 years ago
5

PLEASE HELP!!! I ONLY HAVE A FEW MINUTES!!

Mathematics
1 answer:
TEA [102]2 years ago
6 0

Answer:

y=0.25x-6

Step-by-step explanation:

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An exam consists of 50 multiple choice questions. Based on how much you studied, for
Ann [662]

Using the Central Limit Theorem, it is found that the sampling distribution of the sample proportion of the 50 questions on which you get the correct is approximately normal, with mean of 0.7 and standard error of 0.0648.

<h3>What does the Central Limit Theorem state?</h3>

It states that for a proportion p in a sample of size n, the sampling distribution of sample proportion is approximately normal with mean \mu = p and standard deviation s = \sqrt{\frac{p(1 - p)}{n}}, as long as np \geq 10 and n(1 - p) \geq 10.

In this problem, we have that p = 0.7, n = 50, hence the mean and the standard deviation are given as follows:

\mu = p = 0.7

s = \sqrt{\frac{p(1 - p)}{n}} = \sqrt{\frac{0.7(0.3)}{50}} = 0.0648

More can be learned about the Central Limit Theorem at brainly.com/question/24663213

5 0
2 years ago
What is 67/100 as a decimal representation and a percent representation
bogdanovich [222]

Answer:

Decimal Representation: 0.67

Percent Representation: 67%

4 0
3 years ago
Read 2 more answers
Help, I need it by 9:00 am
olya-2409 [2.1K]
<h3>Answer:</h3>

Option B. because all rational numbers are integers.

Have a look at the above venn diagram showing the relationship between - Real, irrational, rational, integers, whole and natural numbers.

\longrightarrow{\green{Option \:  B.}}⟶Option  \: B.

4 0
2 years ago
What is 857/45 as a mixed number? having trouble with sixth grade math.
barxatty [35]
<span>857/45 as a mixed number is </span>19 2/45
4 0
3 years ago
The average amount of time that students use computers at a university computer center is 36 minutes with a standard deviation o
frosja888 [35]

Answer:

2119 students use the computer for more than 40 minutes. This number is higher than the threshold estabilished of 2000, so yes, the computer center should purchase the new computers.

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore 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 pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

\mu = 36, \sigma = 5

The first step to solve this question is finding the proportion of students which use the computer more than 40 minutes, which is 1 subtracted by the pvalue of Z when X = 40. So

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

Z = \frac{40 - 36}{5}

Z = 0.8

Z = 0.8 has a pvalue of 0.7881.

1 - 0.7881 = 0.2119

So 21.19% of the students use the computer for longer than 40 minutes.

Out of 10000

0.2119*10000 = 2119

2119 students use the computer for more than 40 minutes. This number is higher than the threshold estabilished of 2000, so yes, the computer center should purchase the new computers.

8 0
3 years ago
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