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mr_godi [17]
3 years ago
7

Number 7 on my integrated geometry test

Mathematics
1 answer:
kap26 [50]3 years ago
6 0

Answer:

B. diameter

Step-by-step explanation:

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The screen on Brianna's new cellphone is 2.85 cm long. What mixed number represents this length?
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Answer:

2 17/20

Step-by-step explanation:

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Simplify 9 - {x - (7 + x)}.
LekaFEV [45]

9 - (x - 7 - x) \\ 9 - ( - 7) \\ 9 + 7 \\  = 16
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Use the simple interest formula to find the ending balance.
spayn [35]
5000 * 0.05 * 3.5 = 875

5000 + 875 = 5875

answer: <span>ending balance $5,875</span>
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4 years ago
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Since the area under the normal curve within two standard deviations of the mean is 0.95, the area under the normal curve that c
alukav5142 [94]

Answer:

X \sim N (\mu ,\sigma)

And for this case we know this condition:

P(\mu-2\sigma

By the complement rule we know that:

P(X< \mu -2\sigma \cup X>\mu +2\sigma) = 1-0.95=0.05

But since the distribution is symmetrical we know that:

P(X\mu +2\sigma) = 0.025

So then the statement for this case is FALSE.

b. False

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

Solution to the problem

For this case if we define the random variable of interest X and we know that this random variable follows a normal distribution:

X \sim N (\mu ,\sigma)

And for this case we know this condition:

P(\mu-2\sigma

By the complement rule we know that:

P(X< \mu -2\sigma \cup X>\mu +2\sigma) = 1-0.95=0.05

But since the distribution is symmetrical we know that:

P(X\mu +2\sigma) = 0.025

So then the statement for this case is FALSE.

b. False

5 0
4 years ago
What are the degrees of freedom for Student's t distribution when the sample size is 19? d.F. = Use the Student's t distribution
liq [111]

Answer:

a) 18

b) 2.101

Step-by-step explanation:

a) What are the degrees of freedom for Student's t distribution when the sample size is 19?

Degrees of freedom = n - 1

Where n = sample size

= 19 - 1

= 18

b) Use the Student's t distribution to find tc for a 0.95 confidence level when the sample is 19.

(Round your answer to three decimal places.)

We would be determining these using the t distribution table

1 - 0.95 = 0.05 ( for two tailed)

Or

0.05/2 = 0.025(one - tailed)

Hence, the tc(test score) for a 0.95 confidence level when the sample is 19 is 2.101

7 0
3 years ago
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