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DochEvi [55]
3 years ago
9

What is the slope of the line graphed above​

Mathematics
2 answers:
erma4kov [3.2K]3 years ago
3 0
Answer:
B

Explanation:
Slope = rise/run = -2/2 = -1
USPshnik [31]3 years ago
3 0

Answer:

m=-1/2

Step-by-step explanation:

use points: (-2,2),(2,0)

solve for slope: y2-y1/x2-x1

0-2/2+2=-2/4=-1/2

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Write the expression n ⋅ n ⋅ p ⋅ p ⋅ r ⋅ r ⋅ r using exponents.
hoa [83]

Answer:

n² p²r³

Step-by-step explanation:

n ⋅ n ⋅ p ⋅ p ⋅ r ⋅ r ⋅ r

Write using exponents

There are 2 n terms

n²⋅ p ⋅ p ⋅ r ⋅ r ⋅ r

There are 2 p terms

n² p²⋅ r ⋅ r ⋅ r

There are 3 r terms

n² p²r³

5 0
1 year ago
36°<br> Diagram NOT<br> accurately drawn<br> Work out the size of angle y.
vovikov84 [41]
The angles in a isosceles triangle add up to 180 degrees.
so 180 degrees take away 36 degrees = 144 degrees.
base angles in a isoceles triangle are equal so 144 divided by 2 = 72 degrees.
so to find out y you do 360 - 72 = 288 degrees
therefore y = 288degrees
3 0
3 years ago
Read 2 more answers
The distance between 5,-8 and 5,7 hurry please
solmaris [256]

Answer:

it should be 15

Step-by-step explanation:

5 0
2 years ago
Read 2 more answers
The perimeter of a rectangle is 72 in. The base is 3 times the height. Find the area of the rectangle
Dovator [93]

Answer:

243 in²

Step-by-step explanation:

Step 1. Calculate the<em> base and height </em>of the rectangle

We have two conditions:

(1)   2b+ 2h = 72 in

(2)           b = 3h               Substitute in (1)

2(3h) + 2h = 72               Remove parentheses

   6h + 2h = 72               Combine like terms

           8h = 72               Divide by 6

             h = 9 in             Substitute in(2)

             b = 3 × 9

             b = 27 in

Step 2. Calculate the area of the rectangle

A = bh

A = 27 × 9

A = 243 in²

The area of the rectangle is 243 in².

7 0
3 years ago
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
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