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allsm [11]
3 years ago
6

Find the population mean or sample mean as indicated. Sample: 17, 11, 8, 12, 22

Mathematics
1 answer:
Dominik [7]3 years ago
6 0

Answer:

mean:12

Step-by-step explanation:

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Which is the graph of f(x)=2(3)x​
larisa86 [58]

Answer:A 1st graph

Step-by-step explanation:

4 0
3 years ago
Compare your estimate to the exact quotient. Does your quotient seem reasonable? Explain your answer. 9 1/3 ÷ 2 3/4
dezoksy [38]

Answer:

(9(3)+1)/3÷(2(4)+3)/4

28/3÷11/4

28/3×4/11

112/33

3.393

3 0
3 years ago
Which of the following represents the factorization of the trinomal below
seraphim [82]
C because you factor it out
5 0
3 years ago
A researcher wants to see if birds that build larger nests lay larger eggs. He selects two random samples of nests: one of small
True [87]

Answer:

95% confidence interval for the difference between the average mass of eggs in small and large nest is between a lower limit of 0.81 and an upper limit of 2.39.

Step-by-step explanation:

Confidence interval is given by mean +/- margin of error (E)

Eggs from small nest

Sample size (n1) = 60

Mean = 37.2

Sample variance = 24.7

Eggs from large nest

Sample size (n2) = 159

Mean = 35.6

Sample variance = 39

Pooled variance = [(60-1)24.7 + (159-1)39] ÷ (60+159-2) = 7619.3 ÷ 217 = 35.11

Standard deviation = sqrt(pooled variance) = sqrt(35.11) = 5.93

Difference in mean = 37.2 - 35.6 = 1.6

Degree of freedom = n1+n2 - 2 = 60+159-2 = 217

Confidence level = 95%

Critical value (t) corresponding to 217 degrees of freedom and 95% confidence level is 1.97132

E = t×sd/√(n1+n2) = 1.97132×5.93/√219 = 0.79

Lower limit = mean - E = 1.6 - 0.79 = 0.81

Upper limit = mean + E = 1.6 + 0.79 = 2.39

95% confidence interval for the difference in average mass is (0.81, 2.39)

3 0
3 years ago
What is the answer please help
maria [59]

Answer:

25

Step-by-step explanation:

x^2=24^2+7^2

    =576+49

   =625

x=25

8 0
4 years ago
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