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Maurinko [17]
2 years ago
12

The high temperature one day was -1 °F. The low temperature was -5 °F. What was the difference between the high and low temperat

ures for the day? and why​
Mathematics
1 answer:
insens350 [35]2 years ago
5 0

Answer:

The difference is 4 degrees

Step-by-step explanation:

If its difficult for you to subtract negatives just switch them to a positive 1 and 5 now subtract 1 from 5=4  and because the measure being used is degrees the final answer would be 4 degrees.

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16/20 = 48/60

960 = 960

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An industrial expert claims that the average useful lifetime of a typical car transimssion which comes with ten years warranty i
grin007 [14]

Answer:

Step-by-step explanation:

The question is incomplete. The missing information is the group of answer choices. The group of answer choices are

a) none of the above

b) Data provides sufficient evidence, at 1% significance level, to reject the expert's claim. In addition the p-value (or the observed significance level) is equal to P( T < -3.281).

c) Data provides insufficient evidence, at 1% significance level, to support the expert's claim. In addition the p-value (or the observed significance level) is equal to P( Z > 2.896).

d) Data provides sufficient evidence, at 1% significance level, to support the expert's claim. In addition the p-value (or the observed significance level) is equal to P( T >3.355).

e) Data provides insufficient evidence, at 1% significance level, to support the researcher's claim. In addition the p-value (or the observed significance level) is equal to P(Z > 2.896).

Solution:

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 ≥ 10

For the alternative hypothesis,

µ < 10

This is a left tailed test.

Since the number of samples is small and the population standard deviation is not given, the distribution is a student's t.

Since n = 9,

Degrees of freedom, df = n - 1 = 9 - 1 = 8

t = (x - µ)/(s/√n)

Where

x = sample mean = 13.5

µ = population mean = 10

s = samples standard deviation = 3.2

t = (13.5 - 10)/(3.2/√9) = 3.28

Since α = 0.01, the critical value is determined from the t distribution table. Recall that this is a left tailed test. Therefore, we would find the critical value corresponding to 1 - α and reject the null hypothesis if the test statistic is less than the negative of the table value.

1 - α = 1 - 0.01 = 0.99

The negative critical value is - 2.896

Since - 3.28 is lesser than - 2.896, then we would reject the null hypothesis.

By using probability value,

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

p = 0.0056

Level of significance = 1%

Since alpha, 0.01 > than the p value, 0.0056, then we would reject the null hypothesis. Therefore, At a 1% level of significance, the sample data showed significant evidence that the average useful lifetime of a typical car transimssion which comes with ten years warranty is significantly less than 10 years

The correct option is

a) none of the above

8 0
3 years ago
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