Answer
It’s the first one I guess
Since it’s obviously given in the sequence
Hope it’s correct
Sorry if wrong :’(
What we know:
Football field is 100 yards in length
End zones are 10 yards each in length
Perimeter between pylons is 306 2/3 yards
What we need to find:
a. Perimeter and area of one end zone
b. Perimeter and area of without end zones
c. Perimeter and area of playing field with end zones
First we need to find the measurements of the field between pylons using the perimeter of 306 2/3 yards. We already know the length is 100 yards so we need to find width (w).
P=2l + 2w
306 2/3=2 (100) + 2w
306 2/3= 200 + 2w
300 2/3-200=200-200+2w
106 2/3=2w
106 2/3/2=2/2w
53 1/3=w
a. Perimeter=2 (10)+2 (53 1/3)=126 2/3 yards
Area=10×53 1/3=533 1/3 yd²
b. Perimeter=2 (100)+2 (53 1/3)=306 2/3 yards
Area=100×53 1/3=5333 1/3 yd²
c. Perimeter=2 (120) + 2 (53 1/3)=346 2/3 yards
Area=120×53 1/3=6400 yd²
Well assuming that this would be a typical triangle, and not a right angle one, knowing that the sum of all sides adds up to 180 degrees, simply add all of the expressions and one value and make it equal to 180, and then solve for x.
(6x-1) + (X+14) + 20 = 180
6x - 1 + X + 14 = 160
7x - 1 + 14 = 160
7x + 13 = 160
7x = 147
X = 21.
Now solve for the angles by plugging in X.
A = 6x - 1 = 6(21) - 1 = 125 degrees
C = X + 14 = (21) + 14 = 35 degrees.
I believe these are the solutions.
Answer:
129
Step-by-step explanation:
Considering the survey to be representative, you can simply multiply the share of students <em>p</em> preferring “Track & Field” with the whole school population at the same time to estimate the number of such students in the whole school.
First we need to find the relative share <em>p</em> of such answers in the study by dividing it by the sum of answers, assuming that the table is complete for that random sample:
<em>p</em> = 4/(8 + 5 + 4) = 4/17
Then for the whole school we get 550 <em>p</em> ≈ 129.4
