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erastovalidia [21]
3 years ago
12

If i am 51 years old in december 2002, how old am i in december 2017

Mathematics
2 answers:
ira [324]3 years ago
8 0
I would be 66 years old
nadya68 [22]3 years ago
4 0

Answer:

You would be 66 years old

Step-by-step explanation:

First you have to subtract 2017 - 2002

Then, you use that answer and subtract that, and you will get 66

2017 - 2002

= 15

51 - 15

= 66

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( 16,856.70/ X) = 1.3362
tigry1 [53]

Answer:

x=12615.4019

Step-by-step explanation:

16856.70x=1.3362

So:  x = 16856.7/1.3362

x=12615.4019

5 0
3 years ago
A forest cover 43,000 acres. A survey finds that 0.2% of the forest is old growth tress. How many acres of old growth trees are
Sveta_85 [38]

Given:

Total forest area = 43,000

Old growth trees forest = 0.2%

To find:

The area of the old growth trees.

Solution:

We have,

Total forest area = 43,000

Old growth trees forest = 0.2%

Area of the old growth trees = 0.2% of 43,000

                                               = \dfrac{0.2}{100}\times 43000

                                               = 86

Therefore, the area of the old growth trees is 86 acres.

6 0
3 years ago
Which equation in standard form has a graph that passes through the point (-4,-3) and has a slope of 3/8?
Sidana [21]

Answer:

y = 3/8x - 1.5

Step-by-step explanation:

5 0
3 years ago
A filtration process removes a random proportion of particulates in water to which it is applied. Suppose that a sample of water
Doss [256]

Answer:

E(Y)=\frac{1}{25}

Step-by-step explanation:

Let's start defining the random variables for this exercise :

X_{1}: '' The proportion of the particulates that are removed by the first pass ''

X_{2}: '' The proportion of what remains after the first pass that is removed by the second pass ''

Y: '' The proportion of the original particulates that remain in the sample after two passes ''

We know the relation between the random variables :

Y=(1-X_{1})(1-X_{2})

We also assume that X_{1} and X_{2} are independent random variables with common pdf.

The probability density function for both variables is f(x)=4x^{3} for 0 and f(x)=0 otherwise.

The first step to solve this exercise is to find the expected value for X_{1} and X_{2}.

Because the variables have the same pdf we write :

E(X_{1})= E(X_{2})=E(X)

Using the pdf to calculate the expected value we write :

E(X)=\int\limits^a_b {xf(x)} \, dx

Where a= ∞ and b= - ∞ (because we integrate in the whole range of the random variable). In this case, we will integrate between 0 and 1 ⇒

Using the pdf we calculate the expected value :

E(X)=\int\limits^1_0 {x4x^{3}} \, dx=\int\limits^1_0 {4x^{4}} \, dx=\frac{4}{5}

⇒ E(X)=E(X_{1})=E(X_{2})=\frac{4}{5}

Now we need to use some expected value properties in the expression of Y ⇒

Y=(1-X_{1})(1-X_{2}) ⇒

Y=1-X_{2}-X_{1}+X_{1}X_{2}

Applying the expected value properties (linearity and expected value of a constant) ⇒

E(Y)=E(1)-E(X_{2})-E(X_{1})+E(X_{1}X_{2})

Using that X_{1} and X_{2} have the same expected value E(X) and given that X_{1} and X_{2} are independent random variables we can write E(X_{1}X_{2})=E(X_{1})E(X_{2})   ⇒

E(Y)=E(1)-E(X)-E(X)+E(X_{1})E(X_{2}) ⇒

E(Y)=E(1)-2E(X)+[E(X)]^{2}

Using the value of E(X) calculated :

E(Y)=1-2(\frac{4}{5})+(\frac{4}{5})^{2}=\frac{1}{25}

E(Y)=\frac{1}{25}

We find that the expected value of the variable Y is E(Y)=\frac{1}{25}

3 0
3 years ago
<img src="https://tex.z-dn.net/?f=2m%20%2B%207%20%3D%209" id="TexFormula1" title="2m + 7 = 9" alt="2m + 7 = 9" align="absmiddle"
shtirl [24]
9-7 is equal to 2
2/2 is equal to 1
1=m
8 0
3 years ago
Read 2 more answers
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