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aleksley [76]
3 years ago
6

Al, Bert, and Rob determined that the sum of their ages is 62. Bert is twice as old as Rob, who is 10 years older than Al. How o

ld is each?
Mathematics
1 answer:
elena-14-01-66 [18.8K]3 years ago
3 0

Answer: AI is 8, Rob is 18, and Bert is 36

Step-by-step explanation:

Ok, so the sum of their age is 62. Bert is x2 of Rob. Also someone is 10 years older than AI, lets assume Rob is the guy that is 10 years older. x+y+10+(z times 2)=62. So lets solve it. 62-10=52. 52=x+y+(z times 2). Now if you think about it x and y is the same, because the 10 is already minused. So it's now x*2+z*2=52, if you assume it now, z is x+10. So now it is 52=x*2+(x+10)x2. Now we break out the brackets so its x*2+x*2+20=52. Lets simplify it. x*4+20=52, lets swift the 20 the the other side. x*4=52-20, x*4=32. 32/4=8 so 8 is x. Now we know AI is 8, Rob is 18, and Bert is 36!

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Lelechka [254]
Answer: c)1 solution

y=3x+2
y=4+2x
put the value of either y in another
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7 0
3 years ago
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kondaur [170]

Answer:

7 units

Step-by-step explanation:

The distance between the points (2,−3) and (−5,−3) is 7 units.

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In 2002, there were 972 students enrolled at Oakview High School. Since then, the number of students has increased by 1.5% each
Katarina [22]

Answer:

The number of student increased each year by 986.58.

Step-by-step explanation:

Given : In 2002, there were 972 students enrolled at Oakview High School.

To find : The number of students has increased by 1.5% each year ?

Solution :

Using the exponential growth function,

P(t)=P_o(1+r)^t

Here, P_o=972 is the initial population

r=1.5%=0.015 is the rate

t=1 time

P(t)=972(1+0.015)^1

P(t)=972(1.015)

P(t)=986.58

i.e. the number of student increased each year by 986.58.

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3 years ago
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From a large number of actuarial exam scores, a random sample of scores is selected, and it is found that of these are passing s
Mnenie [13.5K]

<u>Supposing 60 out of 100 scores are passing scores</u>, the 95% confidence interval for the proportion of all scores that are passing is (0.5, 0.7).

  • The lower limit is 0.5.
  • The upper limit is 0.7.

In a sample with a number n of people surveyed with a probability of a success of \pi, and a confidence level of \alpha, we have the following confidence interval of proportions.

\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}

In which

z is the z-score that has a p-value of \frac{1+\alpha}{2}.

60 out of 100 scores are passing scores, hence n = 100, \pi = \frac{60}{100} = 0.6

95% confidence level

So \alpha = 0.95, z is the value of Z that has a p-value of \frac{1+0.95}{2} = 0.975, so z = 1.96.

The lower limit of this interval is:

\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.6 - 1.96\sqrt{\frac{0.6(0.4)}{100}} = 0.5

The upper limit of this interval is:

\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.6 + 1.96\sqrt{\frac{0.6(0.4)}{100}} = 0.7

The 95% confidence interval for the proportion of all scores that are passing is (0.5, 0.7).

  • The lower limit is 0.5.
  • The upper limit is 0.7.

A similar problem is given at brainly.com/question/16807970

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