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4 years ago

Find the slope between (7,1) and (4,8).

Mathematics

2 answers:

stealth61 [152]4 years ago

4 0

Step-by-step explanation:

8-1/4-7 = 7/-3 this the answer I guess

Ivanshal [37]4 years ago

3 0

Answer:

-2.33

Step-by-step explanation:

1-8 =-7

7-4=3

-7 divided by 3 = -2.33

You might be interested in

Determine whether the given segments have the same length. Drag and drop each number

s344n2d4d5 [400]

Answer:

Step-by-step explanation:

Length of a segment between two points $(x_1,y_1)$ and $(x_2,y_2)$ is given by the formula,

$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

Coordinates of extreme ends of BC and EF are,

B(1, 4), C(3, -2), E(-3, -2) and F(-1, 1)

Length of BC = $\sqrt{(1-3)^2+(4+2)^2}$

= $\sqrt{4+36}$

= $2\sqrt{10}$ ≈ 6.32 units

Length of EF = $\sqrt{(-3+1)^2+(-2-1)^2}$

= $\sqrt{4+9}$

= $\sqrt{13}$ ≈ 3.61 units

The length of BC is 6.32. The length of EF is 3.61.So, BC and EF don't have the same length.

6 0

3 years ago

!!!!!!!!!!!! Help!!!!!!!!!

Lisa [10]

1) D is your answer.

Isolate the y:

2x = 3y - 5

first add 5 from both sides

2x (+5) = 3y - 5 (+5)

Next, divide 3 from both sides to isolate the y

(2x + 5)/3 = (3y)/3

y = (2x + 5)/3

2) None of the above is your answer

Plug in points for each equation. None of them work for more than 1 or 2 points. (they are not true for all points) ∴ None of the above is your answer

hope this helps

8 0

4 years ago

If the probability of being hospitalized during a year is 0.15 find the probability that no one in the family of 4 will be hospi

Goryan [66]

The probability that no one in the family of 4 will be hospitalized during a year is 0.52

Step-by-step explanation:

The probability for a person to be hospitalized during one year is

$p(h)=0.15$

As a consequence, the probability for a person of NOT being hospitalized during one year is

$p(h^{-1})=1-0.15=0.85$

If we have 4 people in the family, the probability of each of them of NOT being hospitalized during the year is independent from the probability of the others - therefore, these are independent events. This means that the total probability of the 4 people of not being hospitalized during a year is:

$p(h^{-1})^4=(0.85)^4=0.52$