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Assoli18 [71]
2 years ago
12

Question 1(Multiple Choice Worth 5 points)

Mathematics
1 answer:
ioda2 years ago
7 0
15/31 is the answer
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Analyze the data sets below. Which of the following statements are true? Select all that apply.
PtichkaEL [24]

Answer:

C and D

Step-by-step explanation:

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carmen participated in a read-a-thon.MR.cole pledge $4.00 per book and gave carmen $44.00.How many books did carmen read
finlep [7]

Answer:

11 books

Step-by-step explanation:

because if she gets $4 per book and gets $44 in total then that means she has read 11 books because 4 times 11 is 44.

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3 years ago
How can I use models to help me divide??
Ivan
You can use models to help you divide by making the number of models as your divisor. For example let's say I am dividing 8 ÷ 2. So you would make 8 circles (doesn't have to be circle it can be whatever like squares) and then make groups of 2 until you run out of circles. Then how many groups there are is your answer. 8÷2=4. Hope I helped!
4 0
2 years ago
What is the area of the circle to the nearest tenth of a unit?
kodGreya [7K]

Answer:

15.2

Step-by-step explanation:

PiRsquared

Pi x 2.2²

5 0
3 years ago
Researchers fed mice a specific amount of Dieldrin, a poisonous pesticide, and studied their nervous systems to find out why Die
Elodia [21]

Answer:

Step-by-step explanation:

Part A

Mean = (2.2 + 2.4 + 2.5 + 2.5 + 2.6 + 2.7)/6 = 2.48

Standard deviation = √(summation(x - mean)²/n

n = 6

Summation(x - mean)² = (2.2 - 2.48)^2 + (2.4 - 2.48)^2 + (2.5 - 2.48)^2 + (2.5 - 2.48)^2 + (2.6 - 2.48)^2 + (2.7 - 2.48)^2 = 0.1484

Standard deviation = √(0.1484/6

s = 0.16

Standard error = s/√n = 0.16/√6 = 0.065

Part B

Confidence interval is written as sample mean ± margin of error

Margin of error = z × s/√n

Since sample size is small and population standard deviation is unknown, z for 98% confidence level would be the t score from the student t distribution table. Degree of freedom = n - 1 = 6 - 1 = 5

Therefore, z = 3.365

Margin of error = 3.365 × 0.16/√6 = 0.22

Confidence interval is 2.48 ± 0.22

Part C

We would set up the hypothesis test. This is a test of a single population mean since we are dealing with mean

For the null hypothesis,

H0: µ = 2.3

For the alternative hypothesis,

H1: µ > 2.3

This is a right tailed test

Since the number of samples is small and no population standard deviation is given, the distribution is a student's t.

Since n = 6

Degrees of freedom, df = n - 1 = 6 - 1 = 5

t = (x - µ)/(s/√n)

Where

x = sample mean = 2.48

µ = population mean = 2.3

s = samples standard deviation = 0.16

t = (2.48 - 2.3)/(0.16/√6) = 2.76

We would determine the p value using the t test calculator. It becomes

p = 0.02

Assuming significance level, alpha = 0.05.

Since alpha, 0.05 > than the p value, 0.02, then we would reject the null hypothesis. Therefore, At a 5% level of significance, the sample data showed significant evidence that the mean absolute refractory period for all mice when subjected to the same treatment increased.

6 0
3 years ago
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