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2 years ago

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why is it that when you have a power to a power situation, why do you multiply the exponents ? create an example as part of your

explanation

1 answer:

loris [4]2 years ago

6 0

Step-by-step explanation:

For example, (2^4) is 2x2x2x2 or 16

(2^4)^3 Would be 16^3 or 2x2x2x2 3 times so 12 times in a row which is 4x3

Because exponents are for repeating multiplication they multiply when used together

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Please Help! ASAP! WIll give Brainliest

goblinko [34]

Answer:

x/2 = 12.5/5

5 · x = 12.5 · 2

5x = 25

5x / 5 = 25 / 5

x = 5

Step-by-step explanation:

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3 years ago

What is the mean absolute deviation of the following set of data?<br> 10,4,12,4,2,10,10,6

WITCHER [35]

Answer:

4 and 2

Step-by-step explanation:

10,4,12,4,2,10,10,6

4 and 2

4+2=8

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2 years ago

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Laura is baking a cake. The recipe says that she has to mix 64 grams of chocolate powder to the flour. Laura knows that 1 cup of

pickupchik [31]

The answer is <span>She went over by one over six of a cup
</span>
<span>1 cup of that particular chocolate powder has a mass of 128 grams: 1c = 128g

</span><span>64 grams of chocolate powder is x cups.
1c = 128g
x = 64g

1c : 128 g = x : 64g
x = </span>1c : 128 g * 64 g
x = 0.5 c
64 grams of chocolate powder is 0.5 cups = 1/2 cups

<span>She added two over three of a cup of chocolate powder: 2/3 cups

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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

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

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

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

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

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3 years ago

In one revolution a circle covers a distance equal to its

DIA [1.3K]

Answer:

c

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c

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cc

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6 0

3 years ago

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