For a given confidence level, t ? df is larger than z ? . Explain how t ∗ df being slightly larger than z ∗ affects the width of the confidence interval.

The width of confidence interval is a function of the margin of error. If the critical value of t(t*) is slightly larger than the critical value of z(z*), then the width of the confidence interval will be larger.

The margin of error is the product of the critical value and the standard error. Therefore, given the same standard error value, the value of the margin of error will increases based on the value of the critical value.

Since, t* is slightly larger than z*, then the confidence interval, t will be wider.

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Step-by-step explanation:

The width of the CI is directly proportional to critical value. When t* is greater than z value, the t value would then cause the margin of error to be larger and this will in turn cause the width of the confidence interval to be larger.

Greater t*df than z* gives us a bigger margin of error. This would in turn give bigger width of confidence interval. t distribution has greater width confidence interval compared to z distribution.

d) by the cross product property, ab^2 = bc multiplied by ad

step-by-step explanation:

we are given

in the given triangle abc

angle a is 90° and segment ad is perpendicular to segment bc

we can see that

triangle(abc) and triangle(abd) are similar because they have common side and same angle (90 degree)

so, triangles(abc) and abd are similar

since, they are similar

so, the ratio of their sides must be equal

we get

now, we can cross multiply

and we get

so, this could be step to prove it

answer: i could not see part the question so ill explain how to get the answer

step-by-step explanation: for instance the distance she travele d is 10 and she traveled the same amout each mile so do 10 divided by how many miles she drove each time and you ge your answer.