Answer:
probability of selecting a student who plays a sport but does not watch rugby out of the people who play a sport.
Step-by-step explanation:
"Find the probability that a student chosen at random from those who play a sport does not watch rugby."
90-15= 75 students either play a sport OR watch rugby
65+71-75=
136-75=
61 people play a sport AND watches rugby
14 people play a sport and DOES NOT watch rugby
is the probability of selecting a student who plays a sport but does not watch rugby out of the people who play a sport.
Answer:
We need to solve for the 4th side
4th side base = 75.5 -60.5 = 10 feet
4th side height = 16
4th side LENGTH^2 = 10^2 + 16^2
4th side = sq root (356) = 18.8679622641
Trapezoid Area = [(sum of the bases) / 2 ] * height
Trapezoid Area = [(136)/2] * 16
Trapezoid Area = 1,088 square feet, which is MUCH smaller
than 73,084
Step-by-step explanation:
I'm going to go with C.
The inequality isn't greater than OR EQUAL TO which marks out B and D.
The inequality is showing that it is greater than -21 meaning that the shaded area should be towards the positive which in turn marks out A.
Answer:
Step-by-step explanation:
The mentioned relationship for the weight, in pounds, of the kitten with respect to time, in weeks, is

Weight of the kitten after 10 weeks

pounds
This modeled equation is based on the observation of the early age of a kitten where the kitten is in its growth period, but in the early stage the growth rate in the weight of the kitten was the same but the growth of any living beings continues till the adult stage. So, after some time, in real life situation, this weekly change in weight will become zero, So, this model is not suitable to measure the weight of the kitten over the larger time period.
Here, t= 10 weeks is nearby the observed time period, so the linearly modeled equation can be used to predict the weight.
Hence, the weight of the kitten after 10 weeks is 16.5 pounds.
The intersection between the curves are
3, 0
0, 3
The volume of the solids is obtained by
V = ∫ π [ (4 - (y-1)²)² - (3 - y)²] dy with limits from 0 to 3
The volume is
V = 108π/5 or 67.86