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2 years ago

Thad attempted 25 free throws and made 17 of them while Amy attempted 30 and made 20. Who shot the higher percentage?

Mathematics

1 answer:

ahrayia [7]2 years ago

3 0

Answer: Thad-1.471

Amy-1.500

SO AMY SHOT WITH A HIGHER PERCENTAGE

Step-by-step explanation: brainliest would be appreciated

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HELP PLEASE!!!!! 15 PTS AND BRAINLEST!!!

Schach [20]

The formula for volume is Length x width x height.

Bottom: 18 * 12 * 8 = 1728 cubic centimeters.

Middle: 12 * 8 * 6 = 576 cubic centimeters.

Top = 4 *4*4 = 64 cubic centimeters.

Total: 1728 + 576 + 64 = 2368 cubic centimeters.

8 0

3 years ago

Plese help me its timed

Nadusha1986 [10]

Answer:

(3,2) I think

Step-by-step explanation:

X= 3

Y=2

3 +2 =5

3-2=1

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2 years ago

Solve, then check algebraically and graphically. 7x+14= 56

kari74 [83]

Answer:

x = 6

Step-by-step explanation:

7 x = 56-14

7 x = 42

7 x/7 to remain with x only = 42

then x = 6

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3 years ago

What of the following functions is graphed below

leva [86]

The correct answer is C

5 0

2 years ago

Water is added to a cylindrical tank of radius 5 m and height of 10 m at a rate of 100 L/min. Find the rate of change of the wat

nirvana33 [79]

Answer:

$V = \pi r^2 h$

For this case we know that $r=5m$ represent the radius, $h = 10m$ the height and the rate given is:

$\frac{dV}{dt}= \frac{100 L}{min}$

$Q = 100 \frac{L}{min} *\frac{1m^3}{1000L}= 0.1 \frac{m^3}{min}$

And replacing we got:

$\frac{dh}{dt}=\frac{0.1 m^3/min}{\pi (5m)^2}= 0.0012732 \frac{m}{min}$

And that represent $0.127 \frac{cm}{min}$

Step-by-step explanation:

For a tank similar to a cylinder the volume is given by:

$V = \pi r^2 h$

For this case we know that $r=5m$ represent the radius, $h = 10m$ the height and the rate given is:

$\frac{dV}{dt}= \frac{100 L}{min}$

For this case we want to find the rate of change of the water level when h =6m so then we can derivate the formula for the volume and we got:

$\frac{dV}{dt}= \pi r^2 \frac{dh}{dt}$

And solving for $\frac{dh}{dt}$ we got:

$\frac{dh}{dt}= \frac{\frac{dV}{dt}}{\pi r^2}$

We need to convert the rate given into m^3/min and we got:

$Q = 100 \frac{L}{min} *\frac{1m^3}{1000L}= 0.1 \frac{m^3}{min}$

And replacing we got:

$\frac{dh}{dt}=\frac{0.1 m^3/min}{\pi (5m)^2}= 0.0012732 \frac{m}{min}$

And that represent $0.127 \frac{cm}{min}$

5 0

3 years ago