Hello from MrBillDoesMath
Answer:
[email protected] = - sqrt(7)/ 4
which is choice B
Discussion:
This problem can be solved by drawing triangles and looking at ratios of sides or by using the trig identity:
([email protected])^2 + (sin2)^2 = 1
If [email protected] = 3/4
, the
([email protected])^2 + (3/4)^2 = 1 => (subtract (3/4)^2 from both sides)
([email protected])^2 = 1 - (3/4)^2 = 1 - 9/16 = 7/16
So...... taking the square root of both sides gives
[email protected] = +\- sqrt(7)/ sqrt(16) = +\- sqrt(7)/4
But is [email protected] positive or negative? We are told that @ is in the second quadrant and cos(@) is negative in this quadrant, so our answer must be negative
[email protected] = - sqrt(7)/ 4
which is choice B
Thank you,
Mr. B
It would be 20.15 because the 65% of 31 is 20.15
Answer:
F
Step-by-step explanation:
F
zero
Anytime you have zero as a possible answer, you have to consider it carefully. Part of the whole number system is 0. They go up from there. No fraction is a whole number. No decimal is a whole number except those that are equal to a whole number.
The rest are all decimals so they are not whole numbers. Note I just noticed that the other numbers have periods after the choice. There are other whole numbers there if that is the case.
C D E and F are all whole numbers if that is a period after their choice letters.
Answer:
a. We reject the null hypothesis at the significance level of 0.05
b. The p-value is zero for practical applications
c. (-0.0225, -0.0375)
Step-by-step explanation:
Let the bottles from machine 1 be the first population and the bottles from machine 2 be the second population.
Then we have 
, 
, 
and 
, 
, 
. The pooled estimate is given by
a. We want to test 
vs 
(two-tailed alternative).
The test statistic is 
and the observed value is 
. T has a Student's t distribution with 20 + 25 - 2 = 43 df.
The rejection region is given by RR = {t | t < -2.0167 or t > 2.0167} where -2.0167 and 2.0167 are the 2.5th and 97.5th quantiles of the Student's t distribution with 43 df respectively. Because the observed value 
falls inside RR, we reject the null hypothesis at the significance level of 0.05
b. The p-value for this test is given by 
0 (4.359564e-10) because we have a two-tailed alternative. Here T has a t distribution with 43 df.
c. The 95% confidence interval for the true mean difference is given by (if the samples are independent)
, i.e.,
where 
is the 2.5th quantile of the t distribution with (25+20-2) = 43 degrees of freedom. So
, i.e.,
(-0.0225, -0.0375)
Answer:
Step-by-step explanation:
The sum of 83 and a number would be as an expression, where 
is "said number".
