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3 years ago

Select the correct slope and y-intercept for the following eqnation y=-3x+4

Mathematics

2 answers:

Papessa [141]3 years ago

8 0

y-intercept: 4 happy to help

chubhunter [2.5K]3 years ago

6 0

Inside the equation: y = mx + b, m = slope, and b = y-intercept.

With this knowledge, we will look at the problem:

y = -3x + 4

-3 is in the place of m, which is the slope

slope: -3

4 is in the place of b, which is the y-intercept

y-intercept: 4

hope this helps

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A student who correctly answered 60 questions a test received a score of 75%. How many questions were on the test

Iteru [2.4K]

It would be 45 you take 60 x 75%

7 0

3 years ago

PLEASE HELP<br> find the vertex of f(x)=x^2+2x+3 (make sure you show your work)

kotegsom [21]

Answer:

(- 1, 2)

Step-by-step explanation:

Given a quadratic in standard form y = ax² + bx + c : a ≠ 0

Then the x- coordinate of the vertex is

$x_{vertex}$ = - $\frac{b}{2a}$

f(x) = x² + 2x + 3 ← is in standard form

with a = 1 and b = 2, hence

$x_{vertex}$ = - $\frac{2}{2}$ = - 1

Substitute x = - 1 into f(x) for corresponding value of y

f(- 1) = (- 1)² + 2(- 1) + 3 = 1 - 2 + 3 = 2

vertex = (- 1, 2 )

4 0

3 years ago

Find the length of the missing side

inn [45]

<h3>Answer: 9</h3>

================================================

Work Shown:

a = unknown = leg #1

b = 12 = leg #2

c = 15 = hypotenuse

Plug those values into the pythagorean theorem and solve for 'a'

a^2 + b^2 = c^2

a^2 + (12)^2 = (15)^2 .... substitution

a^2 + 144 = 225

a^2 = 225 - 144 ... subtracting 144 from both sides

a^2 = 81

a = sqrt(81) .... applying square root to both sides

a = 9

7 0

3 years ago

How does three digit division works

ludmilkaskok [199]

Just divide as you normally would but 1 or 2 numbers at a time

8 0

3 years ago

Read 2 more answers

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