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2 years ago

10

What is the expression for "the sum of 8 times a number and two"?

1 answer:

den301095 [7]2 years ago

7 0

The last one is the answer

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Simplify by factoring. What are the excluded values? Choose all that apply.

makvit [3.9K]

Answer:

x = -2

x = 8

Step-by-step explanation:

Excluded values are the ones which make the denominator zero

3x² + x - 10

3x² + 6x - 5x - 10

3x(x + 2) - 5(x + 2)

(x + 2)(3x - 5)

x² - 6x - 16

x² - 8x + 2x - 16

x(x - 8) + 2(x - 8)

(x - 8)(x + 2)

[(x + 2)(3x - 5)] ÷ [(x - 8)(x + 2)]

(3x - 5)/(x - 8)

So excluded values are 8, -2

6 0

3 years ago

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

Leon measured the distance each of his four toy cars rolled across the floor. Which car measurement that has a 4 in the thousand

Kobotan [32]

0.714 and so does 0.224 has 4 in the thousandths place, but in shortest distance I believe it is 0.714

8 0

3 years ago

The number of tourists at the beach per weekend in July was 55,000, In November, the number of tourists per weekend was 18,000.

Ganezh [65]

Answer:

The percentage change from July to November is 67.27 %

Step-by-step explanation:

Given as :

The number of tourists at the beach per weekend in the month of July = $x_1$ = 55,000

The number of tourists at the beach per weekend in the month of November = $x_2$ = 18,000

Let the percentage change from July to November = A %

Or, % decrease change = $\dfrac{\tetxtrm new value - \textrm old value}{\textrm old value}$ × 100

So , A % = $\dfrac{\tetxtrm [tex]x_2$ - \textrm $x_1$ }{\textrm $x_1$ }[/tex] × 100

or, A % = $\frac{55000 - 18000}{55000}$ × 100

Or, A % = $\frac{37000}{55000}$ × 100

Or, A = 67.27 %

So percentage change between two months = 67.27 %

Hence The percentage change from July to November is 67.27 % Answer

6 0

3 years ago

Thirty times the square of a non-zero number is equal to 8 times the number. What is the number?

astra-53 [7]

Hello here is a solution :

4 0

3 years ago