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4 years ago

Right now, Simran's father is three times as old as Simran. Five years ago, he was four times as

Mathematics

2 answers:

Molodets [167]4 years ago

4 0

Answer: The required age of Simran is 15 years and her father is 45 years.

Step-by-step explanation: Given that right now, Simran's father is three times as old as Simran and five years ago, he was four times as old as Simran.

We are to find the present ages of Simran and her father.

Let x and y represents the present ages of Simran and her father respectively.

Then, according to the given information, we have

$y=3x~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(i)\\\\y-5=4(x-4)~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(ii)$

Substituting the value of y from equation (i) in equation (ii), we get

$y-5=4(x-5)\\\\\Rightarrow 3x-5=4x-20\\\\\Rightarrow 4x-3x=20-5\\\\\Rightarrow x=15.$

From equation (i), we get

$y=3\times15=45.$

Thus, the required age of Simran is 15 years and her father is 45 years.

Illusion [34]4 years ago

3 0

Hello,

Let's f the age of the father
s the age of Simran

f=3*s
f-5=4*(s-5)
==>3s-5=4s-20
==>s=15
f=3*15=45

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in the year 2001, a person bought a new car for $22000. for each consecutive year after that, the value of the car depreciated b

igomit [66]

Answer:

To the nearest hundred dollars, the car will be worth $17,900 by 2005

Step-by-step explanation:

Firstly, we need to write the depreciation equation

We have this as:

V = I(1 - r)^t

V is the present value which is what we want to calculate

I is the initial value, the amount the cad was bought which is $22,000

r is the rate of change which is 5% = 5/100 = 0.05

t is the time difference which is 2005-2001 = 4

Substituting all these into the depreciation equation, we have it that

V = 22,000(1 -0.05)^4

V = $17,919.1375

To the nearest hundred dollars, that would be;

$17,900

7 0

3 years ago

A random variableX= {0, 1, 2, 3, ...} has cumulative distribution function.a) Calculate the probability that 3 ≤X≤ 5.b) Find the

olchik [2.2K]

Answer:

a) P ( 3 ≤X≤ 5 ) = 0.02619

b) E(X) = 1

Step-by-step explanation:

Given:

- The CDF of a random variable X = { 0 , 1 , 2 , 3 , .... } is given as:

$F(X) = P ( X =< x) = 1 - \frac{1}{(x+1)*(x+2)}$

Find:

a.Calculate the probability that 3 ≤X≤ 5

b) Find the expected value of X, E(X), using the fact that. (Hint: You will have to evaluate an infinite sum, but that will be easy to do if you notice that

Solution:

- The CDF gives the probability of (X < x) for any value of x. So to compute the P ( 3 ≤X≤ 5 ) we will set the limits.

$F(X) = P ( 3=$

- The Expected Value can be determined by sum to infinity of CDF:

E(X) = Σ ( 1 - F(X) )

$E(X) = \frac{1}{(x+1)*(x+2)} = \frac{1}{(x+1)} - \frac{1}{(x+2)} \\\\= \frac{1}{(1)} - \frac{1}{(2)}\\\\= \frac{1}{(2)} - \frac{1}{(3)} \\\\= \frac{1}{(3)} - \frac{1}{(4)}\\\\= ............................................\\\\= \frac{1}{(n)} - \frac{1}{(n+1)}\\\\= \frac{1}{(n+1)} - \frac{1}{(n+ 2)}$