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3 years ago

Solve the Quadratic Equation.

Mathematics

1 answer:

vichka [17]3 years ago

7 0

Answer:

x = ± 2i

Step-by-step explanation:

The equation has no real roots, gut has complex roots

Given

x² = - 4 ( take the square root of both sides )

x = ± $\sqrt{-4}$

= ± $\sqrt{4(-1)}$

[ Note that $\sqrt{-1}$ = i ]

= ± $\sqrt{4}$ × $\sqrt{-1}$ = ± 2i

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WILL GIVE BRAINLIEST<br> 20 POINTS

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Answer:

Step-by-step explanation:

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3 years ago

Which of the following shows the sum of the polynomials −3x2+3y2+4x and 4x2−4y+2? A. −3x2+3y2+4x+4x2−4y+2 B. −x2+3y2+4x−4y+2 C.

Fed [463]

-3x² + 3y² + 4x + 4x² - 4y + 2

Add like terms.

x² + 3y² + 4x - 4y + 2

So the answer is D.

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4 years ago

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The probability of winning a game is 40%. How many times should you expect to win if you play each number of times? 5 times

Sophie [7]

Answer:

twice

Step-by-step explanation:

to find expected value multiply probablility and number of events

0.40 times 5 = 2.0

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3 years ago

Refer to Exercise 3.122. If it takes approximately ten minutes to serve each customer, find the mean and variance of the total s

garri49 [273]

Answer

a. The expected total service time for customers = 70 minutes

b. The variance for the total service time = 700 minutes

c. It is not likely that the total service time will exceed 2.5 hours

Step-by-step explanation:

This question is incomplete. I will give the complete version below and proceed with my solution.

Refer to Exercise 3.122. If it takes approximately ten minutes to serve each customer, find the mean and variance of the total service time for customers arriving during a 1-hour period. (Assume that a sufficient number of servers are available so that no customer must wait for service.) Is it likely that the total service time will exceed 2.5 hours?

Reference

Customers arrive at a checkout counter in a department store according to a Poisson distribution at an average of seven per hour.

From the information supplied, we denote that

$X=$ Customers that arrive within the hour

and since $X$ follows a Poisson distribution with mean $\alpha$ = 7

Therefore,

$E(X)= 7$

$& V(X)=7$

Let $Y =$ the total service time for customers arriving during the 1 hour period.

Now, since it takes approximately ten minutes to serve each customer,

$Y=10X$

For a random variable $X$ and a constant $c$ ,

$E(cX)=cE(X)\\V(cX)=c^2V(X)$

Thus,

$E(Y)=E(10X)=10E(X)=10*7=70\\V(Y)=V(10X)=100V(X)=100*7=700$

Therefore the expected total service time for customers = 70 minutes

and the variance for serving time = 700 minutes

Also, the probability of the distribution $Y$ is,

$p_Y(y)=p_x(\frac{y}{10} )\frac{dx}{dy} =\frac{\alpha^{\frac{y}{10} } }{(\frac{y}{10})! }e^{-\alpha } \frac{1}{10}\\ =\frac{7^{\frac{y}{10} } }{(\frac{y}{10})! }e^{-7 } \frac{1}{10}$

So the probability that the total service time exceeds 2.5 hrs or 150 minutes is,

$P(Y>150)=\sum^{\infty}_{k=150} {p_Y} (k) =\sum^{\infty}_{k=150} \frac{7^{\frac{k}{10} }}{(\frac{k}{10})! }.e^{-7} .\frac{1}{10} \\=\frac{7^{\frac{150}{10} }}{(\frac{150}{10})! } .e^{-7}.\frac{1}{10} =0.002$

0.002 is small enough, and the function $\frac{7^{\frac{k}{10} }}{(\frac{k}{10} )!} .e^{-7}.\frac{1}{10}$ gets even smaller when $k$ increases. Hence the probability that the total service time exceeds 2.5 hours is not likely to happen.

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3 years ago

Find the inverse of each given functions.<br> f(x) = 4x − 12

Paha777 [63]

Answer:

y=1/4x+3

Step-by-step explanation:

To find the inverse, switch x and y or f(x)

f(x)=4x-12

y=4x-12

x=4y-12

Add 12 to both sides

x+12=4y-12+12

x+12=4y

Divide both sides by 4

x+12/4=y

1/4x+12/4=y

1/4x+3=y

y=1/4x+3

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3 years ago

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