If rope costs $3.20 per yard or $0.15 per inch which is the better deal?

Josh attempted 56 free throws during the basketball season. He made 57 of them. How many free throws did he make?

Scorpion4ik [409]

Answer:

He made approximately 32 free throws.

Step-by-step explanation:

Assuming that there's a typo in the question and that he made 57% of his attempted free throws, then we can solve it as shown below:

We can apply a rule of three in order to calculate the number of free throws he made. This is done as follows:

56 free throws -> 100%

x free throws -> 57%

56/x = 100/57

100*x = 57*56

x = 57*56/100 = 31.92

It can also be solved by transforming the percentage in a fraction such as 57% = 57/100 and then multiplying it by the total attempts.

free throws made = 56*57/100 = 31.92

He made approximately 32 free throws.

7 0

2 years ago

19 is 25% of what number?

zepelin [54]

Answer:

19/25 * 100 = 76%

Step-by-step explanation:

3 0

2 years ago

HEEEELLLLP PPPPLLLLLZZZZ T^T

Rashid [163]

Answer:

1st pic: -25<-20

2nd pic: point c

Step-by-step explanation:

7 0

2 years ago

Read 2 more answers

a satellite's escape velocity is 6.5 mi/sec, the radius of the earth is 3960 mi, and the earth's gravitational constant is 32.2

Artyom0805 [142]

Answer:

$23.89x10^{6}$ ft

Step-by-step explanation:

$\frac{V^{2} }{R^{2} } = \frac{2g}{R+h} \\$ This is the formula to find satellite's escape velocity V , where R is earth's radius, h is the satellite's height from the earth surface and g is the earth's gravitational constant.

$R^{2}(R+h) \frac{V^{2} }{R^{2} } =R^{2} (R+h)\frac{2g}{R+h}$ (Multiplying to clear the fractions)

(R+h) $V^{2} =R^{2} 2g$

$R+h=\frac{2R^{2}g }{V^{2} }$

$h= \frac{2R^{2} g}{V^{2} } -R$

Now, we can determine height of satellite from the surface of the earth

by putting the values in above equation

$h= \frac{2 * (3960)^{2}*32.2 }{6.5^{2} } -3960\\h= 23.902 x10^{6} - 3960\\ h=23.89x10^{6} ft$

5 0

2 years ago

Suppose that a large mixing tank initially holds 500 gallons of water in which 50 pounds of salt have been dissolved. Another br

aliya0001 [1]

Answer:

$A=1500-1450e^{-\dfrac{t}{250}}$

Step-by-step explanation:

The large mixing tank initially holds 500 gallons of water in which 50 pounds of salt have been dissolved.

Volume = 500 gallons

Initial Amount of Salt, A(0)=50 pounds

Brine solution with concentration of 2 lb/gal is pumped into the tank at a rate of 3 gal/min

$R_{in}$ =(concentration of salt in inflow)(input rate of brine)

$=(2\frac{lbs}{gal})( 3\frac{gal}{min})\\R_{in}=6\frac{lbs}{min}$

When the solution is well stirred, it is then pumped out at a slower rate of 2 gal/min.

Concentration c(t) of the salt in the tank at time t

Concentration, $C(t)=\dfrac{Amount}{Volume}=\dfrac{A(t)}{500}$

$R_{out}$ =(concentration of salt in outflow)(output rate of brine)

$=(\frac{A(t)}{500})( 2\frac{gal}{min})\\R_{out}=\dfrac{A}{250}$

Now, the rate of change of the amount of salt in the tank

$\dfrac{dA}{dt}=R_{in}-R_{out}$

$\dfrac{dA}{dt}=6-\dfrac{A}{250}$

We solve the resulting differential equation by separation of variables.

$\dfrac{dA}{dt}+\dfrac{A}{250}=6\\$The integrating factor: e^{\int \frac{1}{250}dt} =e^{\frac{t}{250}}\\$Multiplying by the integrating factor all through\\\dfrac{dA}{dt}e^{\frac{t}{250}}+\dfrac{A}{250}e^{\frac{t}{250}}=6e^{\frac{t}{250}}\\(Ae^{\frac{t}{250}})'=6e^{\frac{t}{250}}$

Taking the integral of both sides

$\int(Ae^{\frac{t}{250}})'=\int 6e^{\frac{t}{250}} dt\\Ae^{\frac{t}{250}}=6*250e^{\frac{t}{250}}+C, $(C a constant of integration)\\Ae^{\frac{t}{250}}=1500e^{\frac{t}{250}}+C\\$Divide all through by e^{\frac{t}{250}}\\A(t)=1500+Ce^{-\frac{t}{250}}$

Recall that when t=0, A(t)=50 (our initial condition)

$50=1500+Ce^{-\frac{0}{250}}50=1500+Ce^{0}\\C=-1450\\$Therefore the amount of salt in the tank at any time t is:\\A=1500-1450e^{-\dfrac{t}{250}}$

4 0

2 years ago