Answer:
a) 
And if we convert this interval into % we got (47.615%, 58.385%)
b) For this case since the lower value for the confidence interval is a value lower than 0.5 (50%) we don't have enough evidence to conclude at 10% of significance than most students at PCC own a car
Step-by-step explanation:
We can begin calculating the best estimator for the true proportion of students at PCC who own a car:
Part a
The confidence level is 95% , and the significance level would be given by 
and 
. And the critical values would be given by:
The confidence interval for the true proportion is given by the following formula:
Replacing the info given we got:
And if we convert this interval into % we got (47.615%, 58.385%)
Part b
For this case since the lower value for the confidence interval is a value lower than 0.5 (50%) we don't have enough evidence to conclude at 10% of significance than most students at PCC own a car
960 watt-hpd
5 x 60/6 = 50
50 x 960 = 48,000
Answer is: -->> 48,000
Well, from what I can tell he has 1 50 cent piece, 2 Dimes, and 2 Pennies. Meaning 5 objects in total, considering that it asks for "at least ten cents" that would mean the 2 Dimes and the 1 50 cent piece.
Since that is 3 objects out of the 5, being 3/5ths, that would mean he has a 60% chance or probability for the first draw.
Hopefully that helps. ^ ^
{-Ghostgate-}
Answer:
83.4% or 83%
Step-by-step explanation:
when trying to find a percent, divide the numerator (# of correct answers) by the denominator (# of total questions possible to get correct)
Which would be 16.67 divided by 20 which equals .8335. Move the decimal point over two placements which would be 83.35. I rounded it up to 83.4. If trying to find a whole number not a decimal it would be 84%. Depending on the grading scale, it would be a B.
Answer:
J 1
--------
x^2 -x
Step-by-step explanation:
x+1
----------
x^3-x
Factor out an x in the denominator
x+1
----------
x(x^2-1)
We can factor the terms in the parentheses because it is a difference of squares
x+1
----------
x(x-1) (x+1)
Canceling the x+1 terms
1
----------
x(x-1)
Distribute in the denominator
1
--------
x^2 -x
