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4 years ago

One zero of f(x)=x^3-2x^2-5x+6 is 1 what are other zeros of the function

Mathematics

2 answers:

Alexus [3.1K]4 years ago

8 0

Answer:

-2 and 3

Step-by-step explanation:

Given one zero of f(x)=x^3-2x^2-5x+6 to be 1 this means that x-1 is a factor of the polynomial x³-2x²-5x+6

Before we can get the other two factors. We need to divide the polynomial function by x-1 and factorize the resulting quotient as shown in the attachment.

Quotient Q(x) = x²-x-6

Factorizing this function

x²-3x+2x-6 = 0

x(x-3)+2(x-3) = 0

(x+2)(x-3) = 0

x+2 = 0 and x-3 = 0

x = -2 and x= 3

Therefore the other zeros of the function are -2 and 3

elena-s [515]4 years ago

7 0

Answer:

The other zeros are -2 and 3.

Step-by-step explanation:

As 1 is a zero then (x - 1) is a factor so we divide:

x - 1 )x^3-2x^2-5x+6(x^2-x-6

x^3-x^2

-x^2-5x

-x^2+x

-6x+6

...........

x^2 - x - 6 = 0

(x - 3)(x + 2)

x = -2, 3.

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melamori03 [73]

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4 years ago

Read 2 more answers

Cars arrive at the Wendy's drive-through at a rate of 1 car every 5 minutes between the hours of 11:00 PM and 1:00 AM. on Saturd

mestny [16]

Answer:

1) $P(X = 8) = 0.1033$

$P(X = 9) = 0.0688$

2) Expected number of 200 restaurants in which exactly 8 customers use the drive-through: 20.66

Expected number of 200 restaurants in which exactly 9 customers use the drive-through: 13.76

Step-by-step explanation:

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

In which

x is the number of sucesses

$e = 2.71828$ is the Euler number

$\mu$ is the mean in the given time interval.

Question 1. Use the Poisson distribution to calculate the probability that exactly 8 cars will use the drive-through between 12:00 midnight and 12:30 AM on a Saturday night at Wendy's. Do the same for exactly 9 cars.

Cars arrive at the Wendy's drive-through at a rate of 1 car every 5 minutes between the hours of 11:00 PM and 1:00 AM. on Saturday nights. This means that during 30 minutes, 6 cars expected to arrive. So $\mu = 6$ .

P(X = 8)

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

$P(X = 8) = \frac{e^{-6}*(6)^{8}}{(8)!} = 0.1033$

P(X = 9)

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

$P(X = 9) = \frac{e^{-6}*(6)^{9}}{(9)!} = 0.0688$

Question 2. At how many of the 200 restaurants in the survey would you expect exactly 8 customers to use the drive-through? exactly 9 customers?

There is a 10.33 probability that 8 customers would use the drive through for each restaurant.

So of 200, the expected number is

$E(X) = 200*0.1033 = 20.66$

There is a 6.88 probability that 9 customers would use the drive through for each restaurant.

So of 200, the expected number is

$E(X) = 200*0.0688 = 13.76$

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3 years ago

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Oksi-84 [34.3K]

Answer:

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Step-by-step explanation:

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3 years ago

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Oduvanchick [21]

Answer:

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Step-by-step explanation:

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3 years ago

Read 2 more answers

In the equation y=x/6, what is the value of the constant coefficient 'k?'

zhuklara [117]

Any relation is x and y is said to be a Direct variation if one is a constant multiplier of the other .In other words we can say : when one variable changes the other changes in proportion to the first.

The Direct variation equation are of the type y=k.x

Here k is known as constant of proportionality .It has a fixed value.

The given equation is y= $\frac{k}{6}$

If we compare this equation with y=k.x

Then k= $\frac{1}{6}$

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3 years ago