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4 years ago

5

Belts cost $160 per carton, and shoes cost $465 per carton. If an order comes in for a total of 88 cartons for $19,265, what is

the number of cartons of belts?

1 answer:

yanalaym [24]4 years ago

5 0

Hello! The first thing you want to do is subtract 88 from 71 which gives you 17. If you multiply the belts times 71 that gives you 11,360. So then you have 17 left and all you do is multiply the shoe cost by 17 and that gives you 7,905. If you add 11,360+7,905 that gives you 19,265. So the correct answer would be 71!

I hope this helped!

I am, yours most sincerely,
SuperHelperThingy

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What is the standard form equation of the line shown below?

ch4aika [34]

(-1,1)(1,4)
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3 years ago

If the average(arithmetic mean) of 5 and r is 7 and the average of 3 and s is 3 , what is the average of r and s?

Norma-Jean [14]

Answer:

C

Step-by-step explanation:

Calculate the values of both r and s, that is

$\frac{5+r}{2}$ = 7 ( multiply both sides by 2 )

5 + r = 14 ( subtract 5 from both sides )

r = 9

and

$\frac{3+s}{2}$ = 3 ( multiply both sides by 2 )

3 + s = 6 ( subtract 3 from both sides )

s = 3

Hence

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4 years ago

Initially 100 milligrams of a radioactive substance was present. After 6 hours the mass had decreased by 3%. If the rate of deca

Hitman42 [59]

Answer:

The half-life of the radioactive substance is 135.9 hours.

Step-by-step explanation:

The rate of decay is proportional to the amount of the substance present at time t

This means that the amount of the substance can be modeled by the following differential equation:

$\frac{dQ}{dt} = -rt$

Which has the following solution:

$Q(t) = Q(0)e^{-rt}$

In which Q(t) is the amount after t hours, Q(0) is the initial amount and r is the decay rate.

After 6 hours the mass had decreased by 3%.

This means that $Q(6) = (1-0.03)Q(0) = 0.97Q(0)$ . We use this to find r.

$Q(t) = Q(0)e^{-rt}$

$0.97Q(0) = Q(0)e^{-6r}$

$e^{-6r} = 0.97$

$\ln{e^{-6r}} = \ln{0.97}$

$-6r = \ln{0.97}$

$r = -\frac{\ln{0.97}}{6}$

$r = 0.0051$

So

$Q(t) = Q(0)e^{-0.0051t}$

Determine the half-life of the radioactive substance.

This is t for which Q(t) = 0.5Q(0). So

$Q(t) = Q(0)e^{-0.0051t}$

$0.5Q(0) = Q(0)e^{-0.0051t}$

$e^{-0.0051t} = 0.5$

$\ln{e^{-0.0051t}} = \ln{0.5}$

$-0.0051t = \ln{0.5}$

$t = -\frac{\ln{0.5}}{0.0051}$

$t = 135.9$

The half-life of the radioactive substance is 135.9 hours.

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3 years ago

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3 years ago

Read 2 more answers

At Lesile's party 1/4 of the people had long hair. 1/2 of the people were boys, and 1/4 of the girls had short blonde hair. None

dybincka [34]

Answer:

4 girls have short red hair

Step-by-step explanation:

Total number of guests at the party is 32

1/2 of the guests are boys

The number of girls with long hair is

= 32×1/4

= 8 girls.

The number of girls with short blonde hair is:

= 16×1/4

= 4 girls

Therefore, we apply the formular below to calculate the maximum number of girls with short red hair.

Total number of girls- ( girls with long hair+ girls with short blonde hair)

= 16-(8+4)

=16-12

= 4

Thus, 4 girls have short red hair

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3 years ago