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3 years ago

Terri is rewriting 7 to the 13th power in expanded form. How many factors should she use in her expression?

2 answers:

7 0

13

Because the exponent is really just how many factors you have.
For example 2 to the power 5 would have 5 factors in expanded form.

Greeley [361]3 years ago

5 0

Answer:

Step-by-step explanation:

Factors is a mathematical or statistical term that describes the numbers you multiply to get another number.

Hence, given that Terri should rewrite 7 to the 13th in expanded form,

Then, she should use 13 factors.

All 13 factors are the same thing . . . 7 .

7^13 should be written as 7¹³ .

Therefore: it works like this: 7*7*7*7*7*7*7*7*7*7*7*7*7

Hence, final answer is 13

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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

3 years ago

Solve the system of equations -5x – 2y = 13 and 3x-y = 12 by combining the equations

Ne4ueva [31]

Answer:

x=1 y =15

Step-by-step explanation:

hopefully correct

-5x-2y=13 (i)

3x-y=12

-y=12-3x

y=-12+3x (ii)

substitute ii into i

-5x-2(-12+3x)=13

-5x +24-6x=13

-11x=-11

x=1

substitute x into ii

y=-12+3(1)

y=15

. . x=1 and y =15

4 0

3 years ago

Share 120 in the ratio 1:5

zhuklara [117]

What are you asking here i think it should be 20:100

5 0

3 years ago

What is the y-intercept of the line?<br><br> x 16 24 32<br> y 44 64 84

Shtirlitz [24]

Answer:

The y-intercept of the line is 2

Step-by-step explanation:

You could try to find the slope of the line and then find the intercept with y axis by sloving b = y - ax

But here it seems easy enough once you see that the x values in the table increase by a factor of 8 and the y values in the table increase by a factor of 20.

(By the way, this is enough to give you the slope, because it is 8/20 ). But why bother?

Just extrapolate the table to the left, and you will find the y value where x = 0

x 0 8 16 24 32

y 2 22 44 64 84

So the y-intercept of the line is 2.

Extra:

The full equation of the line is:

y = 8/20x + 2

3 0

3 years ago

Read 2 more answers

Write 5/10 in 3 equivalent forms

son4ous [18]

5/10 in 3 equivalent forms is: 1/2 2/4 4/8

7 0

3 years ago