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3 years ago

The following set of ordered pairs represents a power inverse variation. Find the value of r given that k = 5.

Mathematics

1 answer:

pochemuha3 years ago

4 0

Answer:

r=2

Step-by-step explanation:

y = k x^r is the formula for a direct variation

y = k x^ -r is the formula for a indirect variation

20= 5 (1/2)^ -r

Divide each side by 5

4 = (1/2) ^ -r

Rewriting

2^2 = 2^ -1 ^ -r

2^2 = 2 ^ r

The bases are the same so

2 =r

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A bag contains two six-sided dice: one red, one green. The red die has faces numbered 1, 2, 3, 4, 5, and 6. The green die has fa

gayaneshka [121]

Answer:

the probability the die chosen was green is 0.9

Step-by-step explanation:

Given that:

A bag contains two six-sided dice: one red, one green.

The red die has faces numbered 1, 2, 3, 4, 5, and 6.

The green die has faces numbered 1, 2, 3, 4, 4, and 4.

From above, the probability of obtaining 4 in a single throw of a fair die is:

P (4 | red dice) = $\dfrac{1}{6}$

P (4 | green dice) = $\dfrac{3}{6}$ = $\dfrac{1}{2}$

A die is selected at random and rolled four times.

As the die is selected randomly; the probability of the first die must be equal to the probability of the second die = $\dfrac{1}{2}$

The probability of two 1's and two 4's in the first dice can be calculated as:

= $\begin {pmatrix} \left \begin{array}{c}4\\2\\ \end{array} \right \end {pmatrix} \times \begin {pmatrix} \dfrac{1}{6} \end {pmatrix} ^4$

= $\dfrac{4!}{2!(4-2)!} ( \dfrac{1}{6})^4$

= $\dfrac{4!}{2!(2)!} \times ( \dfrac{1}{6})^4$

= $6 \times ( \dfrac{1}{6})^4$

= $(\dfrac{1}{6})^3$

= $\dfrac{1}{216}$

The probability of two 1's and two 4's in the second dice can be calculated as:

= $\begin {pmatrix} \left \begin{array}{c}4\\2\\ \end{array} \right \end {pmatrix} \times \begin {pmatrix} \dfrac{1}{6} \end {pmatrix} ^2 \times \begin {pmatrix} \dfrac{3}{6} \end {pmatrix} ^2$

= $\dfrac{4!}{2!(2)!} \times ( \dfrac{1}{6})^2 \times ( \dfrac{3}{6})^2$

= $6 \times ( \dfrac{1}{6})^2 \times ( \dfrac{3}{6})^2$

= $( \dfrac{1}{6}) \times ( \dfrac{3}{6})^2$

= $\dfrac{9}{216}$

∴

The probability of two 1's and two 4's in both dies = P( two 1s and two 4s | first dice ) P( first dice ) + P( two 1s and two 4s | second dice ) P( second dice )

The probability of two 1's and two 4's in both die = $\dfrac{1}{216} \times \dfrac{1}{2} + \dfrac{9}{216} \times \dfrac{1}{2}$

The probability of two 1's and two 4's in both die = $\dfrac{1}{432} + \dfrac{1}{48}$

The probability of two 1's and two 4's in both die = $\dfrac{5}{216}$

By applying Bayes Theorem; the probability that the die was green can be calculated as:

P(second die (green) | two 1's and two 4's ) = The probability of two 1's and two 4's | second dice)P (second die) ÷ P(two 1's and two 4's in both die)

P(second die (green) | two 1's and two 4's ) = $\dfrac{\dfrac{1}{2} \times \dfrac{9}{216}}{\dfrac{5}{216}}$

P(second die (green) | two 1's and two 4's ) = $\dfrac{0.5 \times 0.04166666667}{0.02314814815}$

P(second die (green) | two 1's and two 4's ) = 0.9

Thus; the probability the die chosen was green is 0.9

8 0

3 years ago

Ben earns $13 per hour at Starbucks. When he works overtime, he earns time and a half. What is his regular pay for 40 hours of w

zaharov [31]

Answer:

$520

Step-by-step explanation:

Ben earns $13 per hour of regular work

He gets for 40 hours of regular work:

$13*40 = $520

Answer is $520

7 0

4 years ago

Read 2 more answers

The following histogram presents the amounts of silver (in parts per million) found in a sample of rocks. One rectangle from the

gladu [14]

Answer:

The height of the missing rectangle is 0.15

Explanation:

The image attached has the mentioned histogram.

Such histogram presents the relative frequencies for the clases [0,1], [1,2],[2,3], [4,5], and [5,6] Silver in ppm.

Only the rectangle for the class [3,4] is missing.

The height of each rectangle is the relative frequency of the corresponding class.

The relative frequencies must add 1, because each relative frequency is calculated dividing the absolute class frequency by the total number in the sample; hence, the sum of all the relative frequencies is equal to the total absolute class frequencies divided by the same number, yielding 1.

In consequence, you can sum all the known relative frequencies and subtract from 1 to get the missing relative frequency, which is the height of the missing rectangle.

1. Sum of the known relative frequencies:

0.2 + 0.3 + 0.15 + 0.1 + 0.1 = 0.85

2. Missing frequency:

1 - 0.85 = 0.15

3. Conclusion:

The height of the missing rectangle is 0.15

4 0

3 years ago

Owen received $100 for his birthday.he wants to spend 2/10 of his money on a video game.he wants to spend 55/100 of his money on

babymother [125]

An easy way to solve fraction problems like this is to make each fraction have the same denominator. Since he has $100, the denominator will be 100.
2/10=?/100 10*10=100 1*10=20 20/100 of his money he wants to spend on video games
55/100 already has 100 as its denominator, so he wants to spend 55/100 on a skateboard.
3/10=?/100 10*10=100 3*10=100 30/100 is how much he wants to spend on comic books.
Add all of the fractions together to see how much he wants to spend and if he has enough.
20/100+55/100+30/100=105/100
He wants to spend $105, which is $5 more than he has.

8 0

3 years ago

4/9x+1/5x=58 plzhelp

Olenka [21]

Answer:

Solution

x = 90

Step-by-step explanation:

Combine multiplied terms into a single fraction

⋅

\frac{4}{9}x+\frac{1}{5} \cdot x=58

94x+51⋅x=58

⋅

\frac{4x}{9}+\frac{1}{5} \cdot x=58

94x+51⋅x=58

Combine multiplied terms into a single fraction

Multiply by 1

6 0

3 years ago