Answer:
T = 4.541
Step-by-step explanation:
We don't have the standard deviation for the population, so we use the t-distribution.
The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So
df = 4 - 1 = 3
98% confidence interval
Now, we have to find a value of T, which is found looking at the t table, with 3 degrees of freedom(y-axis) and a confidence level of 
. So we have T = 4.541
The answer is approximately 0.11857
You said X = (5y+1) / (y-2)
Multiply each side by (y-2):
X(y-2) = (5y+1)
Eliminate parentheses:
Xy - 2X = 5y + 1
Add 2X to each side:
Xy = 5y + 1 + 2X
Subtract 5y from each side:
(X-5)y = 1 + 2X
Divide each side by (X-5):
<em>y = (1+2X) / (X-5) </em>
Answer:
There were 6 benches in park 1 and 18 benches in park 2.
Step-by-step explanation:
Let x be the no of benches in Park 1 and y in park 2.
Given that there are 12 more benches in park 2 than 1
Writing this in equation form, we have y = x+12 ... i
Next is if 2 benches were transferred from park 2 to park 1, then we have
x+2 in park 1 and y-2 in park 2.
Given that y-2 = twice that of x+2
Or y-2 = 2x+4 ... ii
Rewrite by adding 2 to both sides of equation ii.
y = 2x+6 ... iii
i-iii gives 0 = -x+6
Or x =6
Substitute in i, to have y = 6+12 = 18
Verify:
Original benches 6 and 18.
18 = 6+12 hence I condition is satisfied
18-2 = 2(6+2)
II is also satisfied.
An observation of the Moon was conducted from Friday, November 8, 2013 to Thursday, November 14, 2013. The study of the Moon during this period occurred consistently between the hours of 8 and 9 p.m. EST within the Northern Hemisphere at 37.3346° N, 79.5228° W (Bedford, V.A.). The Moon was noted to be illuminated on the right side and had a dark shadow on the left side indicating a waxing phase. The light region grew over the surface of the Moon with each subsequent night. The first night’s phase was waxing crescent with over 25 percent of the Moon lit up. The next night, the light had grown to cover more of the Moon as it continued through its waxing crescent phase. On November 10th, the Moon exhibited traits of being at first-quarter or …show more content…
An observation of the Moon was conducted from Friday, November 8, 2013 to Thursday, November 14, 2013. The study of the Moon during this period occurred consistently between the hours of 8 and 9 p.m. EST within the Northern Hemisphere at 37.3346° N, 79.5228° W (Bedford, V.A.). The Moon was noted to be illuminated on the right side and had a dark shadow on the left side indicating a waxing phase. The light region grew over the surface of the Moon with each subsequent night. The first night’s phase was waxing crescent with over 25 percent of the Moon lit up. The next night, the light had grown to cover more of the Moon as it continued through its waxing crescent phase. On November 10th, the Moon exhibited traits of being at first-quarter or half-moon status because at least 50 percent of its surface was illuminated. In the following nights, the Moon displayed characteristics of waxing gibbous as the light continued to grow across the moon’s surface from right to left. The Moon was nearing closer to the full moon phase on November 14th as only a very small dark shadow was visible on the left side.
The Moon takes 27.3 days (sidereal month) to complete its actual orbit around the Earth. Like the Sun, the Moon rises in the east and sets in the west each day.
