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notsponge [240]
3 years ago
9

Worth a lot of points

Mathematics
1 answer:
Dima020 [189]3 years ago
3 0
6. Is 6MPH because 12M / 2H
8. Is 40Seconds because 120 / 3
10. Is 113.3 seconds
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What is an equation of the line that passes through the point (3,-1) and (8, 9)?
Evgesh-ka [11]

Answer:

4) y = 2x - 7

Step-by-step explanation:

First we have to find the slope by using the following formula....

let (3, - 1) be (x_{1}, y_{1} ), let (8, 9) be (x_{2} ,y_{2} ), and m be the slope, then....

m = \frac{y_{2} -y_{1} }{x_{2} - x_{1} } = \frac{9 + 1}{8 - 3} = \frac{10}{5} = 2

Now that we know the slope we can find the equation of the function by using point-slope for of the function which look like this.....

y -y_{1} = m(x - x_{1} )\\y + 1 = 2(x - 3)\\y + 1 = 2x - 6\\y = 2x - 7

3 0
3 years ago
Please help! Chapter test tomorrow!!<br>PLEASE i need to show the work, the equations.
Nana76 [90]

Answer:

x=6

Step-by-step explanation:

4x-15=2x-3

+15 +15

4x=2x=12

/2 /2 +12

2x=12

/2 /2

x=6

4 0
3 years ago
Read 2 more answers
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
Circle D circumscribes ABC and ABE. Which statements about the triangles are true?
OleMash [197]
<span>Circle D circumscribes ABC and ABE, The statements that best describe the triangles are: 
</span><span>Statement I: The perpendicular bisectors of ABC intersect at the same point as those of ABE.
Statement II: The distance from C to D is the same as the distance from D to E. Hence, each of them (CD and DE) is a radius of the given circle.
So, the answer is the second option, I and II.

</span>
7 0
3 years ago
Read 2 more answers
Vernon tossed a coin 20 times. The results were 8 head s and 12 tails. What is the experimental probability of tossing heads?
solmaris [256]

Answer:

2/5

Step-by-step explanation:

The experimental probability is

P (heads) = number of heads/ total tosses

                = 8/20

                 = 2/5

8 0
3 years ago
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