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kramer
3 years ago
15

Justin and tyson are beginning an exercise program to train for football season. Justin weighs 170 pounds and hopes to gain 2 po

unds per week. Tyson weighs 215 pounds and hopes to lose 1 pound per week. if the plan works in how many weeks will the boys weigh the same amount. what will the weight be
Mathematics
1 answer:
nataly862011 [7]3 years ago
8 0

At 15 weeks the boys will weigh the same.

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A total of 20 quarters and nickels add up to $4.00. How many nickels are there?
mariarad [96]

Answer:

5 nickels

Step-by-step explanation:

You can setup and solve a system of equations, or you can solve by trial and error until you get the correct answer.

Here is the solution by trial and error.

If all 20 coins are quarters, the value is 20 * $0.25 = $5

That is too much value.

Let's try 16 quarters. 16 quarters are worth 16 * $0.25 = $4.

That is the correct value, but it is only with quarters, and only 16 of them.

We need fewer quarters than 16.

Try 12 quarters: 12 * $0.25 = $3.00

The number of nickels is: 20 - 12 = 8

8 nickels are worth 8 * $0.05 = $0.40

12 quarters and 8 nickels are worth $3.00 + $0.40 = $3.40

There are 20 coins, but the value is too low.

The number of quarters is between 12 and 16.

Try 14 quarters and 6 nickels:

14 * $0.25 + 6 * $0.05 = $3.50 + $0.30 = $3.80

We are closer to $4 but not there yet.

Try 15 quarters and 5 nickels.

15 * $0.25 + 5 * $0.05 = $3.75 + $0.25 = $4

The total value is $4 and there are 20 coins. This is the answer.

15 quarters and 5 nickels works.

Answer: 5 nickels

8 0
3 years ago
Evaluate the expression when m= -2.8 and n=-6.3.<br> 4n - 6m
Alex73 [517]
Answer is 42, hope this helps
5 0
2 years ago
Find the probability of rolling an odd sum or a sum less than 7 when a pair of dice is rolled
Nutka1998 [239]

Answer:

For the pairs that satisfy that the sum is less than 7 we have:

(1,1) (2,1) (1,2) (3,1) (2,2) (1,3) (4,1) (3,2) (2,3) (1,4) (5,1) (4.2) (3,3) (2,4) (1,5)

So we have a total of 15 pairs where the sum is less than 7

And for an odd sum we have the following pairs

(2,1) (1,2) (4,1) (3,2) (2,3) (1,4) (6,2) (5,2) (4,3) (3,4) (2,5) (1,6) (6,3) (5,4) (4,5) (3,6) (6,5) (5,6)

A total of 18 pairs and we have 6 pairs who are in both cases for the sum less than 7 and the sum an odd number that represent the intersection and using the total probability rule we got:

P = \frac{15+18-6}{36}= \frac{27}{33}= 0.818

Step-by-step explanation:

For this case when a pair of dice is rolled we have the following sample pace for the outcomes:

(1,1) (1,2) (1,3) (1,4) (1,5) (1,6)

(2,1) (2,2) (2,3) (2,4) (2,5) (2,6)

(3,1) (3,2) (3,3) (3,4) (3,5) (3,6)

(4,1) (4,2) (4,3) (4,4) (4,5) (4,6)

(5,1) (5,2) (5,3) (5,4) (5,5) (5,6)

(6,1) (6,2) (6,3) (6,4) (6,5) (6,6)

We see 36 possible outcomes and we want to find how many of these pairs we got rolling an odd sum or a sum less than 7

For the pairs that satisfy that the sum is less than 7 we have:

(1,1) (2,1) (1,2) (3,1) (2,2) (1,3) (4,1) (3,2) (2,3) (1,4) (5,1) (4.2) (3,3) (2,4) (1,5)

So we have a total of 15 pairs where the sum is less than 7

And for an odd sum we have the following pairs

(2,1) (1,2) (4,1) (3,2) (2,3) (1,4) (6,2) (5,2) (4,3) (3,4) (2,5) (1,6) (6,3) (5,4) (4,5) (3,6) (6,5) (5,6)

A total of 18 pairs and we have 6 pairs who are in both cases for the sum less than 7 and the sum an odd number that represent the intersection and using the total probability rule we got:

P = \frac{15+18-6}{36}= \frac{27}{33}= 0.818

7 0
3 years ago
What is the unit rate of 15 and 2
Katena32 [7]
Https://www.khanacademy.org/math/in-sixth-grade-math/ratio-and-proportion/unitary-method/v/finding-unit-rates
4 0
3 years ago
Factor 9x^2+66xy+121y^2
Korvikt [17]
Answer: (3x + 11y)^2

Demonstration:

The polynomial is a perfect square trinomial, because:

1) √ [9x^2] = 3x

2) √121y^2] = 11y

3) 66xy = 2 *(3x)(11y)

Then it is factored as a square binomial, being the factored expression:

 [ 3x + 11y]^2

Now you can verify working backwar, i.e expanding the parenthesis.

Remember that the expansion of a square binomial is:

- square of the first term => (3x)^2 = 9x^2
- double product of first term times second term =>2 (3x)(11y) = 66xy
- square of the second term => (11y)^2 = 121y^2

=> [3x + 11y]^2 = 9x^2 + 66xy + 121y^2, which is the original polynomial.

7 0
3 years ago
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