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Daniel [21]
3 years ago
10

A roller coaster at an amusement park rises to a height of 196 feet above the ground. So far, it has risen 36% of this height. T

o estimate the height the roller coaster has risen so far, Jake did the following.
1. Rounded 36% to 40%
2. Rounded 196 to 200
3. 10% of 200 is 20
4. 20 (4) = 80
5. 36% of 196 is about 80

However, Jake’s estimate was too high by almost 10 feet. Why was his estimate off by so much?
He made a computation error.
He rounded the percent up to 40% and the number up to 200.
He made a rounding error, and should have rounded 196 down to 100.
He should have rounded the percent down to 30% and the number down to 190.
Mathematics
1 answer:
sdas [7]3 years ago
3 0

Answer:    B

Step-by-step explanation: I just know

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WILL MARK BRAINLIEST
goblinko [34]

Answer:

y=.12x+20

Step-by-step explanation:

The flat rate for the ride is $20, so that will be added on no matter how far you go. The distance is actually extra information in this problem, as it doesn't actually ask for you to solve, so ignore the 50 miles from the airport. It is $0.12  per mile, meaning that actually does depend on the number of miles travelled, so it must be the x coefficient. This leaves you with y=.12x+20.

7 0
3 years ago
Read 2 more answers
Mary is 21 years old. Her age is 9 years more than 3 times Henry's age. Which equation can be used to determine x, Henry's age i
luda_lava [24]

Answer:

3x + 9 = 21

Step-by-step explanation:

Let Henry's age be x

Mary's age = 9 + 3x

21 = 9 + 3x

Can also be written as 3x + 9 = 21

4 0
2 years ago
A rancher constructs two rectangular horse pastures that share a side, as shown. The pastures are enclosed by 1050 feet of fenci
noname [10]
The complete question in the attached figure

we know that
area each pasture=15000 ft²
area each pasture=x*y
so
15000=x*y-----> equation 1

perimeter two rectangular pastures=1050 ft
perimeter two rectangular pastures=2*[2x+y]+y----> 4x+2y+y---> 4x+3y
so
1050=4x+3y----> divide by 3----> 1050/3=(4/3)x+y
clear variable y
y=350-(4/3)x-----> equation 2

the answer part a) 
y=350-(4/3)x------> <span>it was proved
</span>
Part b)Find the possible lengths and widths of each pasture.
substitute equation 2 in equation 1
15000=x*y-----> 15000=x*[350-(4/3)x]
15000=350x-(4/3)x²----> multiply by 3----> 45000=1050x-4x²
4x²-1050x+45000=0

using a graph tool----> to resolve the second order equation
see the attached figure

the solution is 
x=53.942 ft
x=208.558 ft

15000=x*y-----> y=15000/x

for x=53.94 ft
y=278.09 ft

for x=208.56 ft
y=71.92 ft

the possible lengths and widths of each pasture are
case 1
x=53.94 ft
y=278.09 ft

case 2
x=208.56 ft
y=71.92 ft

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3 years ago
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zloy xaker [14]

Answer:

b

Step-by-step explanation:

i did this b4

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