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3 years ago

Please help me with this!

Mathematics

2 answers:

Anettt [7]3 years ago

8 0

Differences between x-values are 1. Differences between y-values are 2. The y-values are changing 2 times as fast as the x-values, so you are looking for an answer that involves 2x. The appropriate choice is
D. 2x + 3

_____
You can check this result:
for x=1: 2·1 + 3 = 5
for x=4: 2·4 + 3 = 11
These values correspond to those in the table.

erastovalidia [21]3 years ago

6 0

Let's just list the coordinats first.
(1,5)
(2,7)
(3,9)
(4,11)

When you see 3 and 9, you sometimes immediately think it is 3x=y, but you have to look at the other coordinates as well. This can be show in option a, 5x=y. This can not be the anser because 3×5=15≠9. We can eliminate that answer. Option B says that you are adding 2, but when you add 2 to 4, it equals 6 and not 11. The same for C, 4 added to 4 equals 8, not 11.

The one left, D, is the answer. When you put each x value in, you do get the corresponding y value as well. To even make sure, I made a graph with 2x+3=y as the function to see if those values/coordinates would pop up.

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The function f(x) = -5x2 + 13x + 6 represents the height, in meters, of a

AnnZ [28]

Answer:

Step-by-step explanation:

Differentiate the function and set it to zero:

f'(x) = -10x + 13

0 = -10x + 13

x = 1.3 seconds

Plug the x value into f(x):

f(1.3) = -5(1.3)^2 + 13(1.3) + 6 = 14.45 meters

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3 years ago

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taurus [48]

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3 years ago

Suppose that a box contains r red balls and w white balls. Suppose also that balls are drawn from the box one at a time, at rand

dybincka [34]

Answer: Part a) $P(a)=\frac{1}{\binom{r+w}{r}}$

part b) $P(b)=\frac{1}{\binom{r+w}{r}}+\frac{r}{\binom{r+w}{r}}$

Step-by-step explanation:

The probability is calculated as follows:

We have proability of any event E = $P(E)=\frac{Favourablecases}{TotalCases}$

For part a)

Probability that a red ball is drawn in first attempt = $P(E_{1})=\frac{r}{r+w}$

Probability that a red ball is drawn in second attempt= $P(E_{2})=\frac{r-1}{r+w-1}$

Probability that a red ball is drawn in third attempt = $P(E_{3})=\frac{r-2}{r+w-1}$

Generalising this result

Probability that a red ball is drawn in [tex}i^{th}[/tex] attempt = $P(E_{i})=\frac{r-i}{r+w-i}$

Thus the probability that events $E_{1},E_{2}....E_{i}$ occur in succession is

$P(E)=P(E_{1})\times P(E_{2})\times P(E_{3})\times ...$

Thus $P(E)$ = $\frac{r}{r+w}\times \frac{r-1}{r+w-1}\times \frac{r-2}{r+w-2}\times ...\times \frac{1}{w}\\\\P(E)=\frac{r!}{(r+w)!}\times (w-1)!$

Thus our probability becomes

$P(E)=\frac{1}{\binom{r+w}{r}}$

Part b)

The event " r red balls are drawn before 2 whites are drawn" can happen in 2 ways

1) 'r' red balls are drawn before 2 white balls are drawn with probability same as calculated for part a.

2) exactly 1 white ball is drawn in between 'r' draws then a red ball again at $(r+1)^{th}$ draw

We have to calculate probability of part 2 as we have already calculated probability of part 1.

For part 2 we have to figure out how many ways are there to draw a white ball among (r) red balls which is obtained by permutations of 1 white ball among (r) red balls which equals $\binom{r}{r-1}$