All categories

3 years ago

Which one is the right answer.?

Mathematics

1 answer:

alukav5142 [94]3 years ago

3 0

I think it is A but i could be wrong...

You might be interested in

Does this graph represent a function? Why or why not?

Dafna1 [17]

A. Yes because it passes the vertical line test

6 0

3 years ago

Read 2 more answers

At a high school 7 student volunteer for a committtee how many different 5 person committees can be chosen

ki77a [65]

Total number of volunteer students for a committee = 7
Number of committee = 5

Number of ways of selecting the committee from the 7 volunteers = 7C5 = 7!/(5!*2!) = 21 ways.

There are 21 ways in which the 5-committee member can be selected from the 7 volunteers.

5 0

3 years ago

Plz help me with number 13 plz I need your help plz help me I’m begging you plz help me plz it’s an emergency plz I’m begging yo

Goryan [66]

There are 8 points between 1 and 0 so divide 1 by 8.

1 / 8 = 0.125

Each point has a value of 0.125.

You can minus 0.125 from 1 until you reach A.

Or you can multiple 0.125 by the number of points from 0 to A, including A.

1 - 0.125 - 0.125 - 0.125 = 0.625

0.125 * 5 = 0.625

6 0

3 years ago

Read 2 more answers

Winning the jackpot in a particular lottery requires that you select the correct four numbers between 1 and 42 and, in a separa

NARA [144]

Answer:

$\dfrac{1}{3,357,900}$

Step-by-step explanation:

There are

$C^{42}_4=\dfrac{42!}{4!(42-4)!}=\dfrac{42!}{4!\cdot 38!}=\dfrac{38!\cdot 39\cdot 40\cdot 41\cdot 42}{2\cdot 3\cdot 4\cdot 38!}=39\cdot 10\cdot 41\cdot 7=111,930$

different ways to select the four numbers between 1 and 42. Only one of this ways is correct (successful to win).

There are 30 different ways select the single number between 1 and 30. Only one of them is correct.

The probability of winning the jackpot is

$\dfrac{1}{111,930}\cdot \dfrac{1}{30}=\dfrac{1}{3,357,900}$

4 0

3 years ago

Read 2 more answers

If anyone could help?

Trava [24]

Answer:

$y=\frac{5}{7}x$ and $y=\frac{7}{9}x$

Step-by-step explanation:

A simple way to solve this problem is to plug the corresponding x and y into the function. We need only one pair since all the functions are quasi-linear (y=kx) and the increase is proportional.

In $y=\frac{5}{7}x$ when x=3, y=15/4≈2.14

In $y=\frac{3}{5}x$ when x=3, y=1.8

In $y=\frac{7}{9}x$ when x=3, y≈2.33

In $y=\frac{7}{11}x$ when x=3, y≈1.90

We can observe that in two cases, $y=\frac{5}{7}x$ and $y=\frac{7}{9}x$ , y is greater than 2.

5 0

3 years ago