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2 years ago

Three years ago Tom was twice as old as Jean, and in two years the sum of their ages will be 28 years. Find their present ages.

Mathematics

2 answers:

natali 33 [55]2 years ago

4 0

Answer:

Tom is 15 years old and Jean is 9 years old

Step-by-step explanation:

Let call:

x: the age of Tom

y: the age of Jean

From the sentence: three years ago Tom was twice as old as Jean, we get the first equation:

(x-3)=2(y-3)

Solving for x:

x=2y-6+3

x=2y-3

From the sentence: in two years the sum of their ages will be 28 years, we get the second equation:

(x+2)+(y+2)=28

Solving:

x+y+4=28

x+y=28-4

x+y=24

Now, we have two equations with two incognites:

equation 1: x=2y-3

equation 2: x+y=24

For finding the values of x and y we need to replace the x = 2y-3 on the second equation and solve for y, as following:

(2y-3)+y=24

2y+y-3=24

3y=27

y=27/3

y=9

Then, we Replace y=9 on the first equation and get:

x=2(9)-3

x=18-3

x=15

Now, we have the values of x and y that satisfy the two formulate equation, so the answer is x equal to 15 and y equal to 9. So, Tom is 15 years old and Jean is 9 years old

lubasha [3.4K]2 years ago

3 0

Answer:

Jean is 9 and Tom is 15.

Step-by-step explanation:

3 years ago, Tom was 12 and Jean was 6, hence Tom was twice as old as Jean.

Since that was their age 3 years ago, they are currently 15 (Tom) and 9 (Jean).

Add 2 years to each of these ages, you get 17 and 11.

17 + 11 = 28

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dem82 [27]

Answer:

This spyware is detected 66% of the time.

Step-by-step explanation:

We have these following percentages:

70% of the time, the spyware arrives via the internet, and in this case, is detected 60% of the time.

30% of the time, the spyware arrives via e-mail, and in this case, is detected 80% of the time.

What percentage of times is this spyware detected

Sum of 60% of 70%(arrived via spyware, detecte) and 80% of 30%(arrive via e-mail, detected). So

$0.6*0.7 + 0.8*0.3 = 0.42 + 0.24 = 0.66$

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2 years ago

Measure and record all eight angles formed between the transversal EF and the parallel lines AB and CD. Place the label of each

almond37 [142]

Answer:

EHC=131.19

EHD=48.81

CHG=48.81

DHG=131.19

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HGB=48.81

AGF=48.81

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Step-by-step explanation:

This is exactly from Plato

6 0

2 years ago

Read 2 more answers

A certain type of thread is manufactured with a mean tensile strength of 78.3 kilograms and a standard deviation of 5.6 kilogram

azamat

Answer:

(a) The variance decreases.

(b) The variance increases.

Step-by-step explanation:

According to the Central Limit Theorem if we have a population with mean μ and standard deviation σ and we take appropriately huge random samples (n ≥ 30) from the population with replacement, then the distribution of the sample mean will be approximately normally distributed.

Then, the mean of the sample mean is given by,

$\mu_{\bar x}=\mu$

And the standard deviation of the sample mean is given by,

$\sigma_{\bar x}=\frac{\sigma}{\sqrt{n}}$

The standard deviation of sample mean is inversely proportional to the sample size, n.

So, if n increases then the standard deviation will decrease and vice-versa.

(a)

The sample size is increased from 64 to 196.

As mentioned above, if the sample size is increased then the standard deviation will decrease.

So, on increasing the value of n from 64 to 196, the standard deviation of the sample mean will decrease.

The standard deviation of the sample mean for n = 64 is:

$\sigma_{\bar x}=\frac{\sigma}{\sqrt{n}}=\frac{5.6}{\sqrt{64}}=0.7$

The standard deviation of the sample mean for n = 196 is:

$\sigma_{\bar x}=\frac{\sigma}{\sqrt{n}}=\frac{5.6}{\sqrt{196}}=0.4$

The standard deviation of the sample mean decreased from 0.7 to 0.4 when n is increased from 64 to 196.

Hence, the variance also decreases.

(b)

If the sample size is decreased then the standard deviation will increase.

So, on decreasing the value of n from 784 to 49, the standard deviation of the sample mean will increase.

The standard deviation of the sample mean for n = 784 is:

$\sigma_{\bar x}=\frac{\sigma}{\sqrt{n}}=\frac{5.6}{\sqrt{784}}=0.2$

The standard deviation of the sample mean for n = 49 is:

$\sigma_{\bar x}=\frac{\sigma}{\sqrt{n}}=\frac{5.6}{\sqrt{49}}=0.8$

The standard deviation of the sample mean increased from 0.2 to 0.8 when n is decreased from 784 to 49.

Hence, the variance also increases.

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2 years ago

find the slope of the line on the graph write your answer as a fraction or a whole number not a mixed number or decimal

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2 years ago

Rosita and Garth are saving up for a vacation, and begin with a certain amount saved each. If Rosita works 5 hours, she will hav

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Treat each (time, money) pair as an (x, y) pair, and get the slope of the line:
For Rosita, (5, 128), (7, 164): m = (y2 - y1)/(x2 - x1) = (164 - 128)/(7 - 5) = 18, implying that she earns $18/hr. The y-intercept is calculated as: y = 18x + b, 128 = 18*5 + b, b = $38, meaning that she started with $38. Rosita's equation is y = 18x + 38.
For Garth, (3, 124), (8, 194): m = (194 - 124)/(8 - 3) = 14. For 124 = 14*3 + b, b = $82. Garth's equation is y = 14x + 82

To find out when they will have saved the same amount, both equations would have the same y-value:
18x + 38 = 14x + 82
4x = 44
x = 11 hours
y = 18*11 + 38 = $236 (alternatively, y = 14*11 + 82 = 236)
This means that Rosita and Garth will have both saved $236 after 11 hours of working.

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