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3 years ago

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Mathematics

1 answer:

BabaBlast [244]3 years ago

5 0

A) .10
b) 40x.35 = 14

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An environment engineer measures the amount ( by weight) of particulate pollution in air samples ( of a certain volume ) collect

Serggg [28]

Answer:

$k = 1$

$P(x > 3y) = \frac{2}{3}$

Step-by-step explanation:

Given

$f \left(x,y \right) = \left{ \begin{array} { l l } { k , } & { 0 \leq x} \leq 2,0 \leq y \leq 1,2 y \leq x } & { \text 0, { elsewhere. } } \end{array} \right.$

Solving (a):

Find k

To solve for k, we use the definition of joint probability function:

$\int\limits^a_b \int\limits^a_b {f(x,y)} \, = 1$

Where

${ 0 \leq x} \leq 2,0 \leq y \leq 1,2 y \leq x }$

Substitute values for the interval of x and y respectively

So, we have:

$\int\limits^2_{0} \int\limits^{x/2}_{0} {k\ dy\ dx} \, = 1$

Isolate k

$k \int\limits^2_{0} \int\limits^{x/2}_{0} {dy\ dx} \, = 1$

Integrate y, leave x:

$k \int\limits^2_{0} y {dx} \, [0,x/2]= 1$

Substitute 0 and x/2 for y

$k \int\limits^2_{0} (x/2 - 0) {dx} \,= 1$

$k \int\limits^2_{0} \frac{x}{2} {dx} \,= 1$

Integrate x

$k * \frac{x^2}{2*2} [0,2]= 1$

$k * \frac{x^2}{4} [0,2]= 1$

Substitute 0 and 2 for x

$k *[ \frac{2^2}{4} - \frac{0^2}{4} ]= 1$

$k *[ \frac{4}{4} - \frac{0}{4} ]= 1$

$k *[ 1-0 ]= 1$

$k *[ 1]= 1$

$k = 1$

Solving (b): $P(x > 3y)$

We have:

$f(x,y) = k$

Where $k = 1$

$f(x,y) = 1$

To find $P(x > 3y)$ , we use:

$\int\limits^a_b \int\limits^a_b {f(x,y)}$

So, we have:

$P(x > 3y) = \int\limits^2_0 \int\limits^{y/3}_0 {f(x,y)} dxdy$

$P(x > 3y) = \int\limits^2_0 \int\limits^{y/3}_0 {1} dxdy$

$P(x > 3y) = \int\limits^2_0 \int\limits^{y/3}_0 dxdy$

Integrate x leave y

$P(x > 3y) = \int\limits^2_0 x [0,y/3]dy$

Substitute 0 and y/3 for x

$P(x > 3y) = \int\limits^2_0 [y/3 - 0]dy$

$P(x > 3y) = \int\limits^2_0 y/3\ dy$

Integrate

$P(x > 3y) = \frac{y^2}{2*3} [0,2]$

$P(x > 3y) = \frac{y^2}{6} [0,2]\\$

Substitute 0 and 2 for y

$P(x > 3y) = \frac{2^2}{6} -\frac{0^2}{6}$

$P(x > 3y) = \frac{4}{6} -\frac{0}{6}$

$P(x > 3y) = \frac{4}{6}$

$P(x > 3y) = \frac{2}{3}$

8 0

2 years ago

the function f(x) = 3,005(1+0.03)^x represents the amount of money in a savings account where x represents time in years. what d

Stolb23 [73]

Answer:

Option d. the initial amount of money placed in the savings account

Step-by-step explanation:

we have

$f(x)=3,005(1+0.03)^{x}$

This is a exponential function of the form

$f(x)=a(1+r)^{x}$

where

a is the initial value

r is the growth rate

(1+r) is the base

x is the number of years

f(x) is the amount of money in a savings account

In this problem we have

a=$3,005

r=0.03=3%

(1+r)=1.03

therefore

3,005 represent the initial value ( the amount of money for the value of x equal to zero)

6 0

3 years ago

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62 is 50% of what number?

Stels [109]

62 is 50% of 124 (62*2=124)

5 0

3 years ago

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What is true about open-end credit? OA It is not typically used for a single purchase but for many purchases over a period of ti

Nitella [24]

Answer:

Option B

Step-by-step explanation:

Open end credit 's biggest example is our credit card

You use credit card and take the money borrow from bank as per your requirement.
If you pay in time your credit score increases else decreases.

Option B is correct

7 0

1 year ago

Points N and R both lie on circle O. Line segment RQ is tangent to the circle at point R.

4vir4ik [10]

Answer:

15.0

Step-by-step explanation:

Let's start by looking at triangle ORQ. Since RQ is tangent to the circle, we know that angle ∠ORQ is 90°. Then, since OR is equivalent to the radius of 5, RQ is 5√3, and side OQ is clearly larger than RQ, we can identify this as a 90-60-30 degree triangle. This makes side OQ have a length of 10, and angle ∠QOR, opposite of the second largest side, has the second largest angle of 60°, leaving ∠OQR with an angle of 30°.

The formula for the chord length is 2r*sin(c/2), with c being the angle between the two points on the circle (in this case, ∠QOR=∠NOR).. Our radius is 5, so the length of chord NR is 2*5*sin(60/2)=5, making our answer 5(ON)+5(OR)+5(RN)

4 0

3 years ago

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