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Ifteen bales of hale were loaded into a truck. Some of the bales weight 150 pounds, and some weighed 100 punds. Hpw many bales o

horrorfan [7]

Answer:

There are 7 bales of 100 pounds and 8 bales of 150 pounds hale

Step-by-step explanation:

Let the amount of bale of 100 pounds be x and the bale of hale of 150pounds be y

If fifteen bales of hale were loaded into a truck, then x+y = 15 ... 1

Also if some of the bales weight 150 pounds, and some weighed 100 pounds with total weight of pounds to be 1900 pounds, then mathematically;

150y + 100x = 1900

Dividing through by 100;

1.5y+x = 19 ..... 2

Solving equation 1 and 2 simultaneously;

x+y = 15 ... 1

1.5y+x = 19 .... 2

Subtracting 1 from 2

y-1.5y = 15-19

-0.5y = -4

Divide both sides by 0.5

-0.5y/-0.5 = -4/-0.5

y = 8

Substituting y = 8 into equation 1, we will have;

x+y = 15

x+8 = 15

x = 15-8

x = 7

Hence there are 7 bales of 100 pounds and 8 bales of 150 pounds hale

6 0

3 years ago

Billy is monitoring the exponential decay of a radioactive compound. He has a sample of the compound in a test tube in his lab.

Over [174]

Solution:

Formula for radioactive Decay is given by

$R_{0}= R(1-\frac{S}{100})^t$

$R_{0}$ = Initial Population

R = Remaining population after time in hours

Rate of Decay = S % per hour

Initial Population = 72 grams

Final population = 15 grams

Rate of Decay = 35 % per hour

Substituting the values to get value of t in hours

$72=15(1-\frac{35}{100})^t\\\\ 4.8= (0.65)^t\\\\ t= -3.64$ →→1 St expression

But taking positive value of t , that is after 3.64 hours the sample of 72 grams decays to 15 grams at the rate of 35 % per hour.

Now , it is also given that, Once the sample reaches a mass of 15 grams, Billy will continually add more of the compound to keep the sample size at a minimum of 15 grams.

Substituting these in Decay Formula

Final Sample = 15 gm

Starting Sample = 15 +k, where k is amount of sample added each time to keep the final sample to 15 grams.

Time is over 3.64 hours i.e new time = 3.64 + t

Rate will remain same i.e 35 % per hour.

$15=(15+k)(1-\frac{35}{100})^{3.64+t}$ →→→ Final expression (Second) , that is inequalities can be used to determine the possible mass of the radioactive sample over time.