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STatiana [176]
3 years ago
14

Gary is 8 years older than his brother. Six years ago Gary was three times as old as his brother. How old are they now?

Mathematics
1 answer:
Yuliya22 [10]3 years ago
8 0

Answer:Gary-18 years brother-10 years

Step-by-step explanation:1st case: brother=1 yrs,gary=3 yrs=>after 6 after: brother =7 yrs gary=9 years

2nd case:......

3rd case:.....

4th case:brother:4 years gary:12 yrs=>after 6 yrs:brother=10 yrs gary=18 yrs=>18-10=8

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I want to know how to round the number 445,778 nearest 10,000
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450,000 is your answer:) hope I helped
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3 years ago
A+5b-c=-20<br> 4a-5b+4c=19<br> -a-5b-5c=2 <br> Solve for a b and c
ludmilkaskok [199]

Answer:a = - 2

b = - 3

c = 3

Step-by-step explanation:

a + 5b - c = - 20 - - - - - - - - - 1

4a - 5b + 4c = 19 - - - - - - - - - 2

-a - 5b - 5c = 2 - - - - - - - - - - - 3

Adding equation 1 and equation 2, it becomes

- 6c = - 18

c = - 18/ - 6 = 3

Multiplying equation 2 by 1 and equation 3 by 4, it becomes

4a - 5b + 4c = 19

-4a - 20b - 20c = 8

Adding both equations, it becomes

- 25b - 16c = 27

- 25b = 27 + 16c = 27 + 16 × 3

- 25b = 75

b = 75/- 25 = - 3

Substituting b = - 3 and c = 3 into equation 1, it becomes

a + 5 × - 3 - 3 = - 20

a - 15 - 3 = - 20

a - 18 = - 20

a = - 20 + 18 = - 2

5 0
3 years ago
What is the answer. Shawn pays $1.75 for a school lunch on days when he doesn't bring his own lunch to school. One week he bough
katen-ka-za [31]

Answer:

C. 3.50

Step-by-step explanation:

He bought a lunch two times, so $1.75*2 = 3.50.

5 0
3 years ago
The prime factorization of 180X²Y? (PLEASE HELP)
Law Incorporation [45]
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Explanation:it’s easier
8 0
2 years ago
A teacher was interested in knowing the amount of physical activity that his students were engaged in daily. He randomly sampled
klasskru [66]

Answer:

The standard error of the mean is 4.5.

Step-by-step explanation:

As we don't know the standard deviation of the population, we can estimate the standard error of the mean from the standard deviation of the sample as:

\sigma_{\bar{x}}\approx\frac{s}{\sqrt{n}}

The sample is [30mins, 40 mins, 60 mins, 80 mins, 20 mins, 85 mins]. The size of the sample is n=6.

The mean of the sample is:

\bar{x}=\frac{1}n} \sum x_i =\frac{30+40+60+80+20+85}{6}=52.5

The standard deviation of the sample is calculated as:

s=\sqrt{\frac{1}{n-1}\sum (x_i-\bar x)^2} \\\\ s=\sqrt{\frac{1}{5}\cdot ((30-52.5)^2+(40-52.5)^2+(60-52.5)^2+(80-52.5)^2+(20-52.5)^2+(85-52.5)^2}\\\\s=\sqrt{\frac{1}{5} *3587.5}=\sqrt{717.5}=26.8

Then, we can calculate the standard error of the mean as:

\sigma_{\bar{x}}\approx\frac{s}{\sqrt{n}}=\frac{26.8}{6}= 4.5

6 0
3 years ago
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