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KATRIN_1 [288]
3 years ago
6

Help please I don’t understand

Mathematics
2 answers:
natta225 [31]3 years ago
7 0
79 I really hope this helps
german3 years ago
6 0

Answer: 79 degrees

Step-by-step explanation:

85 - 6 = 79

Hope this helps!

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A randomly generated list of integers from 1 to 5 is being used to simulate an
elena55 [62]

Answer: it’s 40% if your confused… aka (A.)

Step-by-step explanation:

4 0
3 years ago
Todd’s average score for six tests was 92. If the sum of the scores of two of her tests was 188, then what was her average score
natita [175]

Answer:

91

Step-by-step explanation:

Todd’s average score for six tests = 92.

The sum of two of her test = 188

First, we need to find the total score for the six test. This given below:

Average = sum of all test / number of test

sum of all the test = average x number of test

average score for six tests = 92.

Number of test = 6

Sum of all the Tests = 92 x 6 = 552

Sum of four test = sum of all the test — sum of two test

Sum of four test = 552 — 188 = 364

Now we can solve for the average of the other four test as shown below:

Average of four test = 364/4= 91

3 0
3 years ago
Read 2 more answers
5*10 divided by 30 -10
sp2606 [1]
The answer to your question would be -25/3

As decimal -8.333333
4 0
2 years ago
A researcher took a poll of people in his state. He asked each person if he or she had any pets (Yes or No) and to state his or
Arlecino [84]

Answer:

DF = 2

Step-by-step explanation:

Formula for degree of freedom is;

DF = n - 1

Where n is sample size

In this question, political party affiliation could be Republican, Democrat, or Other. Thus, sample size is 3.

DF = 3 - 1

DF = 2

Thus, degree of freedom value would be 2

8 0
3 years ago
The exam scores for 200 students are normally distributed with a mean of 72 and a standard deviation of 10. Which answer choice
Paha777 [63]

Using the normal distribution, it is found that there are 68 students with scores between 72 and 82.

<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.

In this problem, the mean and the standard deviation are given, respectively, by:

\mu = 72, \sigma = 10

The proportion of students with scores between 72 and 82 is the <u>p-value of Z when X = 82 subtracted by the p-value of Z when X = 72</u>.

X = 82:

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

Z = \frac{82 - 72}{10}

Z = 1

Z = 1 has a p-value of 0.84.

X = 72:

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

Z = \frac{72 - 72}{10}

Z = 0

Z = 0 has a p-value of 0.5.

0.84 - 0.5 = 0.34.

Out of 200 students, the number is given by:

0.34 x 200 = 68 students with scores between 72 and 82.

More can be learned about the normal distribution at brainly.com/question/24663213

#SPJ1

6 0
2 years ago
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