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Tju [1.3M]
3 years ago
8

Justin is saving money to buy a stereo. He has $25 saved in the bank right now. He earns $40 each week delivering newspapers.

Mathematics
1 answer:
mafiozo [28]3 years ago
7 0

Answer:

y = 25 + 40x

Step-by-step explanation:

Let

y = the total amount of money Justin has

x = number of weeks

Amount Justin has in the bank = $25

Amount Justin earns per week = $40

Equation for how much money Justin has (including the amount he has in the bank) in x weeks

the total amount of money Justin has = Amount Justin has in the bank + (Amount Justin earns per week * number of weeks)

y = 25 + (40 * x)

y = 25 + 40x

The equation is

y = 25 + 40x

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Adam had some candy to give to his three children. He first took seven pieces for himself and then evenly divided the rest among
blondinia [14]

Answer : The expression will be

( c - 7 ) / 3

Step-by-step explanation:

Let's say Adam has c pieces of candy

He took ten pieces for himself which means we can write this as c-7

He then divided this amount between his 3 children (c-7)/3

4 0
3 years ago
The experimental probability of a biased coin landing on heads is 0.6
Setler [38]

Answer:

32

Step-by-step explanation:

1-.60=.4 is the probability for tails

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3 years ago
PLEASE ANSWER!
butalik [34]

We just plug in, simplify, and solve.

x/2 + 3x - y

8/2 + 3(8) - 2

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Option C should be your answer.

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4 0
3 years ago
How would i solve a problem like this?
SVETLANKA909090 [29]

The number of combinations of size k that you can make with n items is given by the so-called binomial coefficient,

\dbinom nk = \dfrac{n!}{k!(n-k)!}

• n! is the number of ways of permuting n items.

• (n-k)! is the number of ways of permuting all but k of the n items.

Dividing n! by (n-k)! then gives the number of ways of permuting only k of the total n items.

• k! is the number of ways of permuting k items.

Dividing \frac{n!}{(n-k)!} by k! then removes all those permutations which contain the same items. We call these combinations.

For this problem we only care about counting combinations.

There are

\dbinom 63 = \dfrac{6!}{3!(6-3)!} = 20

ways of selecting any 3 girls from the total 6 girls in the entire group of people.

There are

\dbinom 72 = \dfrac{7!}{2!(7-2)!} = 21

was of selecting any 2 boys from the total 7 boys.

Then there are

\dbinom 63 \dbinom 72 = 20\cdot21 = \boxed{420}

ways of choosing a committee of 5 people consisting of 3 girls and 2 boys.

If the next question were, "What is the probability that a committee of 5 randomly selected people consists of 3 girls and 2 boys?", then you would additionally need to compute the number of ways one can make a committee of 5 people from the total 13, which is

\dbinom{13}5 = \dfrac{13!}{5!(13-5)!} = 1287

Then the probability of selecting such a committee at random is

\dfrac{\binom 63 \binom72}{\binom{13}5} = \dfrac{420}{1287} = \dfrac{140}{429} \approx 0.3263

8 0
2 years ago
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