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3 years ago

Is the graphed function linear?

1 answer:

6 0

Step-by-step explanation: Any function of the form f (x) = m x + b, where m is not equal to 0 is called a linear function. The domain of this function is the set of all real numbers. The range of f is the set of all real numbers. The graph of f is a line with slope m and y intercept b.

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What is 1 1/4 × 4 3/4 =

Luden [163]

First, we need to turn this into an improper fraction. Therefore we have: 1 1/4 = 5/4 and 4 3/4 = 19/4
Now we multiply across and eliminate as needed.
5/4 * 19/4 = 95/16
Finally, we reduce and convert to a mixed fraction.
16 goes into 95, 5 times with a remainder of 15. Therefore, we can conclude that our answer is: 5 15/16.

4 0

3 years ago

In ABC the measure of angle A is 2x+4 , the measure of angle B is 4x+2 , and the measure of angle C is 2x-1. What are the measur

Lynna [10]

(2x+4)+(4x+2)+(2x-1)=180
8x+5=180°
8x=180-5
8x=175
X=21.875
A= 2(21.875)+4=47.75
B=4(21.875)+2 =89.5
C=2(21.875)-1=42.75

5 0

3 years ago

Read 2 more answers

Which expression is equivalent to (73)-2

padilas [110]

Answer:

73*-2 =-146 you have to multiply the 73 to the -2

Step-by-step explanation:

6 0

3 years ago

12. 4(2x - 8) = 3(2 – 3x)<br> Need help to solve it in steps

Reil [10]

8x - 32 = 6 - 9x
17x = 38
x = 38/17

4 0

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 $\binom{r}{r-1}$

Thus the probability becomes $P(E_i)=\frac{\binom{r}{r-1}}{\binom{r+w}{r}}=\frac{r}{\binom{r+w}{r}}$

Thus required probability of case b becomes $P(E)+ P(E_{i})$

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

7 0

3 years ago