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3 years ago

Write the quadratic question f(x)=3x^2-24x+49 in the form of f(x)= a(x-h)^2+k

Mathematics

1 answer:

vredina [299]3 years ago

7 0

Answer:

f(x) = 3(x -4)^2 +1

Step-by-step explanation:

Usually you start by factoring the leading coefficient from the first two terms:

f(x) = 3(x^2 -8x) +49

Now, you add the square of half the x-coefficient inside parentheses, and subtract the same quantity outside.

f(x) = 3(x^2 -8x +(-4)^2) -3(-4)^2 +49

= 3(x -4)^2 -48 +49

f(x) = 3(x -4)^2 +1 . . . . . . . . the vertex form you desire

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2b+8 is the simplify expression

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2 years ago

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Delaney has $28.00 and buys a shirt for 14.00. Delaney used the equation 14x = 28 to model her scenario, where x represents the

maksim [4K]

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2 years ago

Question 4 options: Find the mean and standard deviation for a binomial distribution with 680 trials and a probability of succes

lys-0071 [83]

Answer:

Mean for a binomial distribution = 374

Standard deviation for a binomial distribution = 12.97

Step-by-step explanation:

We are given a binomial distribution with 680 trials and a probability of success of 0.55.

The above situation can be represented through Binomial distribution;

$P(X=r) = \binom{n}{r}p^{r} (1-p)^{n-r} ; x = 0,1,2,3,.....$

where, n = number of trials (samples) taken = 680 trials

r = number of success

p = probability of success which in our question is 0.55

So, it means X ~ $Binom(n=680, p=0.55)$

Now, we have to find the mean and standard deviation of the given binomial distribution.

Mean of Binomial Distribution is given by;

E(X) = n $\times$ p

So, E(X) = 680 $\times$ 0.55 = 374

Standard deviation of Binomial Distribution is given by;

S.D.(X) = $\sqrt{n \times p \times (1-p)}$

= $\sqrt{680 \times 0.55 \times (1-0.55)}$

= $\sqrt{680 \times 0.55 \times 0.45}$ = 12.97

Therefore, Mean and standard deviation for binomial distribution is 374 and 12.97 respectively.

3 0

3 years ago

Solve for x and y: 3/x -1/y =13/10 1/x + 2/y =9/10

harkovskaia [24]

Answer:

(x, y) = (2, 5)

Step-by-step explanation:

I find it easier to solve equations like this by solving for x' = 1/x and y' = 1/y. The equations then become ...

3x' -y' = 13/10

x' +2y' = 9/10

Adding twice the first equation to the second, we get ...

2(3x' -y') +(x' +2y') = 2(13/10) +(9/10)

7x' = 35/10 . . . . . . simplify

x' = 5/10 = 1/2 . . . . divide by 7

Using the first equation to find y', we have ...

y' = 3x' -13/10 = 3(5/10) -13/10 = 2/10 = 1/5

So, the solution is ...

x = 1/x' = 1/(1/2) = 2

y = 1/y' = 1/(1/5) = 5

(x, y) = (2, 5)

_____

The attached graph shows the original equations. There are two points of intersection of the curves, one at (0, 0). Of course, both equations are undefined at that point, so each graph will have a "hole" there.

3 0

3 years ago