All categories

3 years ago

Help with this i dont understand

Mathematics

1 answer:

velikii [3]3 years ago

3 0

Answer:

a) 3/4

b) 3/4

c) 3/4

d) all are 3/4

Step-by-step explanation:

The slope of a line is defined as the change in y divided by the change in x. This is often referred to as "rise over run":

slope = Δy/Δx = "rise"/"run"

The triangles are drawn against the line so that you can use them to identify values of "rise" (vertical change) and "run" (horizontal change) between two points on the line.

The slope of the segment AC is found by dividing the "rise" (BC = 3) by the "run"(AB = 4)

slope of AC = BC/AB = 3/4

The slope of FH is similarly computed. Here, it is made a little more difficult, because you must estimate the value of GH.

slope of FH = GH/FG = 1.5/2 = 0.75 = 3/4

The slope of CE is ...

slope of CE = DE/CD = 6/8 = 3/4

The slopes in a), b), c) are all 3/4.

You can reasonably draw the conclusion that the slope of a line is the same everywhere. A line has constant slope.

You might be interested in

I need help finding the value of x n y. thank you

rewona [7]

Step-by-step explanation:

Here with reference angle 45°

Hypotenuse ( h) = y

Perpendicular (p) = x

base (b) = 5√2

We know

tan 45° = p/b

1 = x / 5√2

Therefore x = 5√2

Now

h = √(5√2)^2 + 5√2)^2

= √ 50 +50

= √100

= 10

Hope it helped

4 0

3 years ago

Five cards are drawn from a standard 52-card playing deck. A gambler has been dealt five cards—two aces, one king, one 3, and on

Nookie1986 [14]

Answer:

The probability that he ends up with a full house is 0.0083.

Step-by-step explanation:

We are given that a gambler has been dealt five cards—two aces, one king, one 3, and one 6. He discards the 3 and the 6 and is dealt two more cards.

We have to find the probability that he ends up with a full house (3 cards of one kind, 2 cards of another kind).

We know that gambler will end up with a full house in two different ways (knowing that he has given two more cards);

If he is given with two kings.
If he is given one king and one ace.

Only in these two situations, he will end up with a full house.

Now, there are three kings and two aces left which means at the time of drawing cards from the deck, the available cards will be 47.

So, the ways in which we can draw two kings from available three kings is given by = $\frac{^{3}C_2 }{^{47}C_2}$ {∵ one king is already there}

= $\frac{3!}{2! \times 1!}\times \frac{2! \times 45!}{47!}$ {∵ $^{n}C_r = \frac{n!}{r! \times (n-r)!}$ }

= $\frac{3}{1081}$ = 0.0028

Similarly, the ways in which one king and one ace can be drawn from available 3 kings and 2 aces is given by = $\frac{^{3}C_1 \times ^{2}C_1 }{^{47}C_2}$