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Alexus [3.1K]
3 years ago
6

(X) = \frac{(X-3)(X+1)}{(X+3)}" alt="F(X) = \frac{(X-3)(X+1)}{(X+3)}" align="absmiddle" class="latex-formula">
What is the:
Domain:
Holes:
VA:
HA:
OA: Roots:
Y-Intercepts:
Mathematics
1 answer:
Nezavi [6.7K]3 years ago
7 0

Answer:

Step-by-step explanation:

Given function is

f(x)=\frac{(x-3)(x+1)}{(x+3)}

Domain of the function: (-∞,-3)∪(-3,∞) x | x ≠ -3

Holes: none. No common factors in numerator and denominator

VA: x = -3 as denominator is (x+3)

HA: No horizontal asymptote because degree of the denominator is less than numerator

OA: Roots are x = 3, -1 [(x-3) and (x+1) are the factors]

Y-Intercept: (0, -1) (by putting x = 0 in the given function)

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Step-by-step explanation:

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5 0
2 years ago
A gasoline tank for a certain car is designed to hold 14.0 gal of gas. Suppose that the variable x = actual capacity of a random
MariettaO [177]

Answer:

A gasoline tank for a certain car is designed to hold 14.0 gal of gas.

\mu = 14

\sigma = 0.2

a)What is the probability that a randomly selected tank will hold at most 13.7 gal?

We are supposed to find P(x \leq 13.7)

z=\frac{x-\mu}{\sigma}

z=\frac{13.7-14}{0.2}

z=-1.5

Refer the z table for p value

p value = 0.0668

So, P(x \leq 13.7)= 0.0668

The probability that a randomly selected tank will hold at most 13.7 gal is  0.0668

b)What is the probability that a randomly selected tank will hold between 13.4 and 14.3 gal?

We are supposed to find P(13.4

z=\frac{x-\mu}{\sigma}

z=\frac{13.4-14}{0.2}

z=-3

P(x<13.4)=P(z<-3)= 0.0013

z=\frac{x-\mu}{\sigma}

z=\frac{14.3-14}{0.2}

z=1.5

P(x<14.3)= 0.9332

P(13.4

The probability that a randomly selected tank will hold between 13.4 and 14.3 gal is 0.9319

(c) If two such tanks are independently selected, what is the probability that both hold at most 14 gal?

We are supposed to find P(x \leq 14)

z=\frac{x-\mu}{\sigma}

z=\frac{14-14}{0.2}

z=0

Refer the z table for p value

p value = 0.5

So, P(x \leq 14)= 0.5

Two such tanks are independently selected, the probability that both hold at most 14 gal = 0.5 * 0.5 = 0.25

Hence If two such tanks are independently selected, the probability that both hold at most 14 gal is 0.25

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HELP PLS which box has the largest IQR
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Answer:

Box B i think.

Step-by-step explanation:

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