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Dima020 [189]
3 years ago
14

The price of pair of shoes is $63.20. The sales tax rate is 4.5 percent. How much sales tax would you pay if you bought these sh

oes
Mathematics
1 answer:
VMariaS [17]3 years ago
7 0

Answer:2.84

Step-by-step explanation: 4.5% of 63.20

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Answer:

T,F,T,F

Step-by-step explanation:

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Yesterday morning at 8 a.m. the temperature was -14˚F. At noon it was 10 degrees warmer. The temperature increase between noon a
OLEGan [10]
The temperature was -14°F at 8 a.m.

At noon is was 10 degrees warmer - the temperature increase was 10°F
-14°F + 10°F = -4°F

The tempertaure increase between noon and 4 p.m. was twice the previous increase in temperature, which was 10°F.
10°F x 2 = 20°F

We want to know the temperature at 4 p.m. so we just add it up to the previous result.
-4°F + 20°F = 16°F

16°F is the correct answer
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3 years ago
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Find a8 of the sequence 10,9.75,9.5,9.25,….
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Answer:

10,9.75,9.5,9.25,9, 8.75 , 8.5, 8.25, 8...

Step-by-step explanation:

Subtract 0.25 from each to find the next number

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3 years ago
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Will mark Brainliest<br><br>1/4 ÷ 7/1 =
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1/4 divided by 7/1 equals 1/28
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2 years ago
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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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