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Stolb23 [73]
2 years ago
12

A is two years older than b who is twice as old as c. If the total of the ages of a, b and c be 27, then how old is b?.

Mathematics
1 answer:
RoseWind [281]2 years ago
8 0
A is 12 years
b is 10 years
c is 5 years
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Which equation does the graph of the systems of equations solve?
Viktor [21]

Answer:

We conclude that option 'C' i.e. 1/3x+1=2x-4 is the correct option.

Step-by-step explanation:

From the given graph, it is clear that the two lines meet at the point (3, 2).

In other words,

  • at x =3, y = 2

Thus,

The point of intersection between two lines is:

(x, y) = (3, 2)

Here:

x = 3 is the value of the x-coordinate

y = 2 is the value of the y-coordinate

Now, checking the equation i.e. 1/3x+1=2x-4 to determine whether the equation contains the correct value of x-coordinate or not.

\frac{1}{3}x+1=2x-4

Subtract 1 from both sides

\frac{1}{3}x+1-1=2x-4-1

Simplify

\frac{1}{3}x=2x-5

subtract 2x from both sides

\frac{1}{3}x-2x=2x-5-2x

Simplify

-\frac{5}{3}x=-5

Multiply both sides by 5

3\left(-\frac{5}{3}x\right)=3\left(-5\right)

Simplify

-5x=-15

Divide both sides by -5

\frac{-5x}{-5}=\frac{-15}{-5}

Simplify

x=3

Therefore, the value of x = 3

Conclusion:

We conclude that option 'C' i.e. 1/3x+1=2x-4 is the correct option.

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Answer:

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Step-by-step explanation:

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The annual Gross Domestic Product (GDP) of a country is the value of all of the goods and services produced in the country durin
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F=5.77X10^9(1.032^(y-1985)), we want to know when this is:

2.2X10^12=5.77X10^9(1.032^(y-1985))

381.2825=1.032^(y-1985)

ln(3,812825)=(y-1985)ln(1.032)

ln(3.812825)/ln(1.032)=y-1985

y=1985+ln(3.812825)/ln(1.032)

y≈2027.5  

So GDP will reach 2.2 trillion during the year 2027.
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3 years ago
Elijah earns $200 for 8 hours of work.What is the rate (in dollars per item)
MAXImum [283]
25 dollars per hour should be the correct answer
6 0
3 years ago
34​% of college students say they use credit cards because of the rewards program. You randomly select 10 college students and a
finlep [7]

Answer:

a) There is a 18.73% probability that exactly two students use credit cards because of the rewards program.

b) There is a 71.62% probability that more than two students use credit cards because of the rewards program.

c) There is a 82% probability that between two and five students, inclusive, use credit cards because of the rewards program.

Step-by-step explanation:

There are only two possible outcomes. Either the student use credit cards because of the rewards program, or they use for other reason. So, we can solve this problem by the binomial distribution.

Binomial probability

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

P(X = x) = C_{n,x}.\pi^{x}.(1-\pi)^{n-x}

In which C_{n,x} is the number of different combinatios of x objects from a set of n elements, given by the following formula.

C_{n,x} = \frac{n!}{x!(n-x)!}

And \pi is the probability of X happening.

In this problem, we have that:

10 student are sampled, so n = 10

34% of college students say they use credit cards because of the rewards program, so \pi = 0.34

(a) exactly​ two

This is P(X = 2).

P(X = x) = C_{n,x}.\pi^{x}.(1-\pi)^{n-x}

P(X = 2) = C_{10,2}.(0.34)^{2}.(0.66)^{8} = 0.1873

There is a 18.73% probability that exactly two students use credit cards because of the rewards program.

(b) more than​ two

This is P(X > 2).

Either a value is larger than two, or it is smaller of equal. The sum of the decimal probabilities must be 1. So:

P(X \leq 2) + P(X > 2) = 1

P(X > 2) = 1 - P(X \leq 2)

In which

P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2)

So

P(X = x) = C_{n,x}.\pi^{x}.(1-\pi)^{n-x}

P(X = 0) = C_{10,0}.(0.34)^{0}.(0.66)^{10} = 0.0157

P(X = 1) = C_{10,1}.(0.34)^{1}.(0.66)^{9} = 0.0808

P(X = 2) = C_{10,2}.(0.34)^{2}.(0.66)^{8} = 0.1873

P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2) = 0.0157 + 0.0808 + 0.1873 = 0.2838

P(X > 2) = 1 - P(X \leq 2) = 1 - 0.2838 = 0.7162

There is a 71.62% probability that more than two students use credit cards because of the rewards program.

(c) between two and five inclusive

This is:

P = P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5)

P(X = x) = C_{n,x}.\pi^{x}.(1-\pi)^{n-x}

P(X = 2) = C_{10,2}.(0.34)^{2}.(0.66)^{8} = 0.1873

P(X = 3) = C_{10,3}.(0.34)^{3}.(0.66)^{7} = 0.2573

P(X = 4) = C_{10,4}.(0.34)^{4}.(0.66)^{6} = 0.2320

P(X = 5) = C_{10,5}.(0.34)^{5}.(0.66)^{5} = 0.1434

P = P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) = 0.1873 + 0.2573 + 0.2320 + 0.1434 = 0.82

There is a 82% probability that between two and five students, inclusive, use credit cards because of the rewards program.

6 0
3 years ago
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