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Sergeeva-Olga [200]
3 years ago
13

Question 6

Mathematics
2 answers:
Paul [167]3 years ago
7 0
9.28 divided by 2.9 is equal too the first answer which is 3.2
77julia77 [94]3 years ago
4 0

Answer:

9.28/2.9=3.2is a required answer

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Malachi has sticks that represent 4 different segment lengths: 3 inches, 4 inches, 5 inches, and 6 inches.
Dahasolnce [82]

Answer:

Create a triangle with the 3-, 4-, and 5-inch sticks and measure the angles to show that one angle is a right angle.

Step-by-step explanation:

6 0
2 years ago
Suppose you received a score of 95 out of 100 on exam 1 . The mean was 79 and the standard deviation was 8 . If your score on ex
Schach [20]

Answer:

You did the same on both exams.

Step-by-step explanation:

To compare both the scores, we need to compute the z scores of both the exams and then compare the values. The formula for z-score is:

<u>Z = (X - μ)/σ</u>

Where X = score obtained

           μ = mean score

           σ = standard deviation

For Exam 1:

Z = (95 - 79)/8

  = 16/8

<u>Z = 2</u>

For Exam 2:

Z = (90 - 60)/15

  = 30/15

<u>Z = 2</u>

<u>The z-scores for both the tests are same hence the third option is correct i.e. </u><u>you did the same on both exams.</u>

8 0
3 years ago
That is, what is the 142nd odd number?​
Kitty [74]

Answer and step-by-step explanation:

I believe it is 287. I am not entirely sure, but I believe that is the answer.

4 0
3 years ago
According to the Pew report, 14.6% of newly married couples in 2008 reported that their spouse was of another race or ethnicity
sweet [91]
Shisiekkd. Wklwlwl kalwow wish is owl am dh Wii’s all wish wishfully usnan2839 7 hessian
3 0
3 years ago
A random sample of 49 statistics examinations was taken. the average score, in the sample, was 84 with a variance of 12.25. the
tester [92]

Since in this case we are only using the variance of the sample and not the variance of the real population, therefore we use the t statistic. The formula for the confidence interval is:

<span>CI = X ± t * s / sqrt(n)                      ---> 1</span>

Where,

X = the sample mean = 84

t = the t score which is obtained in the standard distribution tables at 95% confidence level

s = sample variance = 12.25

n = number of samples = 49

From the table at 95% confidence interval and degrees of freedom of 48 (DOF = n -1), the value of t is around:

t = 1.68

 

Therefore substituting the given values to equation 1:

CI = 84 ± 1.68 * 12.25 / sqrt(49)

CI = 84 ± 2.94

CI = 81.06, 86.94

 

<span>Therefore at 95% confidence level, the scores is from 81 to 87.</span>

6 0
4 years ago
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