An independent-samples t test was conducted to compare high-altitude exercise to exercise at sea-level. Imagine the mean comfort level of participants, rated on a 7-point scale where high values reflect high comfort, at high altitude was 3.5 while the average at sea-level was 5.1. Imagine also that values were calculated for sdifference of 1.1, 2.04 for pooled variance, and 35 for degrees of freedom total. What are the critical cutoffs for this test

-2.03, 2.03

Step-by-step explanation:

The degree of freedom (df) ;

Df total for the two samples = 35

Therefore, obtuaking the critical value using a T-test distribution at 95% confidence level.

(1 - 95%) / 2 = (1 - 0.95) / 2 = 0.05 / 2 = 0.025

Using a T critical value calculator or t table distribution ;

Tcritical, 0.025, df = 35 = 2.03

Left tail = - 2.03

Right = 2.03

Hence, critical cutoff :

(-2.03 ; 2.03)

d

step-by-step explanation:

d

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