16 boards that are 5.7 inches wide each make a total of 91.20 inches.
16
<u>x 5.7
</u> 112
<u>+800
</u> 91.2 inches
Since the deck itself is 94.45 inches deep, we can subtract 91.2 from the number:
94.45
<u>-91.20</u>
3.25 inches.
Finally, we divide 3.25 by 15, since that's the amount of spaces we want and they have to be equal:
3.25/15= .21667
We'll round that to two numbers, so we'll end up with .22.
Your final answer is .22 inches wide for each space.
Answer:
a)We are 95% confident that the average commuting time for route A is between 1.3577 and 4.6423 minutes shorter than the average committing time for rout B.
(b) No, because the confidence internal does not contain —5, which corresponds with an average of 5 minutes shorter for route A.
Step-by-step explanation:
Given:
n_1 = 20
x_1= 40
s_1 = 3
n_2 = 20
x_2= 43
s_2 = 2
d_f = 33.1
c = 95%. 0.95
(a) Determine the t-value by looking in the row starting with degrees of freedom df = 33.1 > 32 and in the column with c = 95% in the Student's t distribution table in the appendix:
t
/2 = 2.037
The margin of error is then:
E = t
/2 *√s_1^2/n_1+s_2^2/n_2
E = 2.037 *√3^2/20+s_2^2/20
= 1.64
The endpoints of the confidence interval for u_1 — u_2 are:
(x_1 — x_2) — E = (40 — 43) — 1.6423 = —3 — 1.6423= —4.6423
(x_1 - x_2) + E = (40 — 43) + 1.6423 = —3 + 1.6423= —1.3577
a)We are 95% confident that the average commuting time for route A is between 1.3577 and 4.6423 minutes shorter than the average committing time for rout B.
(b) No, because the confidence internal does not contain —5, which corresponds with an average of 5 minutes shorter for route A.
Slope intercept form is y = mx + b where m is the slope and b is the y-intercept.
Plugging those in we get:
y = - 5/2x + 1
Answer:
-41
Step-by-step explanation:
g(x) = 3x + 1
g(-14) = 3*(-14) + 1
= -42 + 1
= -41
Answer: 2
Step-by-step explanation: