Answer:
V=576
Step-by-step explanation:
It is D you have subtract them
1. The shape of cross-section is a circle.
2. The face parallel to ABCD is EFGH. Since this is a a rectangular shape,
A = L*H = 12*6 = 72 cm^2
3. The cross-section parallel to ABC is DEF with h = 12 ft, b= 5ft (where h is the height and b is the base of a right angled triangle).
Area, A = 1/2 *b*h = 1/2*5*12 =30 ft^2
4. Plane BDHF is a rectangle shape whose length is the diagonal of ABCD.
Diagonal BD = sqrt (AB^2+BD^2) = sqrt (8^2+7^2) = 10.63 cm.
Perimeter, P = 2(BD+DH) = 2(10.63+6) = 33.26 cm
We are given the following function and want to find the y-intercept:

Just substitute the value x=0 into the equation to find the value of the y-intercept:

Simplify

Simplify again

Since the value of f(x), or y, when x=0 is 66, the y-intercept of this function is 66.
Let me know if you need any clarifications, thanks!
~ Padoru
Answer:
Since the calculated value of z= -1.496 does not fall in the critical region z < -1.645 we conclude that the new program is effective. We fail to reject the null hypothesis .
Step-by-step explanation:
The sample proportion is p2= 7/27= 0.259
and q2= 0.74
The sample size = n= 27
The population proportion = p1= 0.4
q1= 0.6
We formulate the null and alternate hypotheses that the new program is effective
H0: p2> p1 vs Ha: p2 ≤ p1
The test statistic is
z= p2- p1/√ p1q1/n
z= 0.259-0.4/ √0.4*0.6/27
z= -0.141/0.09428
z= -1.496
The significance level ∝ is 0.05
The critical region for one tailed test is z ≤ ± 1.645
Since the calculated value of z= -1.496 does not fall in the critical region z < -1.645 we conclude that the new program is effective. We fail to reject the null hypothesis .