Simplify the following: a. 9 sinA +33 cosec A + 10 sin A - 13 cosec A
1 answer:
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Answer:
This data suggest that there is more variability in low-dose weight gains than in control weight gains.
Step-by-step explanation:
Let be the variance for the population of weight gains for rats given a low dose, and
the variance for the population of weight gains for control rats whose diet did not include the insecticide.
We want to test vs
. We have that the sample standard deviation for
female control rats was
g and for
female low-dose rats was
g. So, we have observed the value
which comes from a F distribution with
degrees of freedom (numerator) and
degrees of freedom (denominator).
As we want carry out a test of hypothesis at the significance level of 0.05, we should find the 95th quantile of the F distribution with 17 and 21 degrees of freedom, this value is 2.1389. The rejection region is given by {F > 2.1389}, because the observed value is 3.3176 > 2.1389, we reject the null hypothesis. So, this data suggest that there is more variability in low-dose weight gains than in control weight gains.
Answer:
the last option:
Step-by-step explanation:
Make sure you have the numerical answer for each of the functional expressions that are shown among the possible solution choices:
f(4) = -14 (what the blue function reads [its y-value] when x is 4)
g(4) = 10 (what the red function reads [its y-value] for x=4)
g(-2) = 4 (y-value of the red function when x is -2)
f(2) = -8 (y-value of the blue function for x = 2)
f(-2) = 4 (y-value of the blue function for x =-2)
use them to compare the options they give you, and the only one that matches is the last option.