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Ksju [112]
4 years ago
8

Teresa is going for a run. she runs for 9.6 miles at a speed of 6 miles per hour. for how many hours does she run?

Mathematics
1 answer:
Lera25 [3.4K]4 years ago
7 0
Speed=6 miles per hour
Distance =9.6 miles
Speed =distance /time
Time=distance/speed=9.6/6=1.6 hours
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What is the slope between the points (-5 7) and (4 -8)?
german

Answer:

15/-9

Step-by-step explanation:

y2-y1/x2-X1

7-(-8) 15

-5-4. -9

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What does a straight line on a distance time graph mean
klasskru [66]
It means the speed (velocity) was constant - the object traveled the same distance per time period throughout the graph.
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Describe the translation from f ( x ) = I x I... to f ( x ) = I x - 1 I + 11.
noname [10]
Horizontal shift 1 unit left and vertical shift 11 units up
6 0
2 years ago
What is equivalent to 3 and 1/4 minus negative 5/8
marysya [2.9K]

Answer:

31/8

Step-by-step explanation:

3/1/4 - (-5/8)

13/4 + 5/8

find Lcm of 4 and 8 = 8

multiply their common factors with numerators.

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=31/8 or 3/7/8 or 3.875

7 0
3 years ago
Read 2 more answers
For parts a and bâ, use technology to estimate the following. âa) The critical value of t for a 90â% confidence interval with df
rjkz [21]

Answer:

a) For the 90% confidence interval the value of \alpha=1-0.9=0.1 and \alpha/2=0.05, with that value we can find the quantile required for the interval in the t distribution with df =3. And we can use the folloiwng excel code: "=T.INV(0.05,3)" and we got:

t_{\alpha/2} =\pm 2.35

b) For the 99% confidence interval the value of \alpha=1-0.99=0.01 and \alpha/2=0.005, with that value we can find the quantile required for the interval in the t distribution with df =106. And we can use the folloiwng excel code: "=T.INV(0.005,106)" and we got:

t_{\alpha/2} =\pm 2.62

Step-by-step explanation:

Previous concepts

The t distribution (Student’s t-distribution) is a "probability distribution that is used to estimate population parameters when the sample size is small (n<30) or when the population variance is unknown".

The shape of the t distribution is determined by its degrees of freedom and when the degrees of freedom increase the t distirbution becomes a normal distribution approximately.  

The degrees of freedom represent "the number of independent observations in a set of data. For example if we estimate a mean score from a single sample, the number of independent observations would be equal to the sample size minus one."

Solution to the problem

Part a

For the 90% confidence interval the value of \alpha=1-0.9=0.1 and \alpha/2=0.05, with that value we can find the quantile required for the interval in the t distribution with df =3. And we can use the folloiwng excel code: "=T.INV(0.05,3)" and we got:

t_{\alpha/2} =\pm 2.35

Part b

For the 99% confidence interval the value of \alpha=1-0.99=0.01 and \alpha/2=0.005, with that value we can find the quantile required for the interval in the t distribution with df =106. And we can use the folloiwng excel code: "=T.INV(0.005,106)" and we got:

t_{\alpha/2} =\pm 2.62

3 0
4 years ago
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