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2 years ago

Without using technology, describe the end behavior of f(x) = −3x4 + 7x2 − 12x + 13.

Mathematics

1 answer:

rodikova [14]2 years ago

8 0

Answer:

11 over 13

Step-by-step explanation:

first, you deal with multiplication;

x = -12 + 14 - 12x + 13

second, deal with addition;

x = -2 - 12x + 13

third, take -12 to the other side then the negative changes to positive;

12x + x = -2 + 13

fourth, add them then simplify;

13x = 11

ans is 11 over 13

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Jose must plant 380 trees. In the past 7 days Jose planted 266 trees. If he continues at this rate, how many more days will it t

Sonja [21]

Answer:

16 days

Step-by-step explanation:

because

380-266=114

114 divided by 7= 16

3 0

2 years ago

The number of flaws in a fiber optic cable follows a Poisson distribution. It is known that the mean number of flaws in 50m of c

boyakko [2]

Answer:

(a) The probability of exactly three flaws in 150 m of cable is 0.21246

(b) The probability of at least two flaws in 100m of cable is 0.69155

(c) The probability of exactly one flaw in the first 50 m of cable, and exactly one flaw in the second 50 m of cable is 0.13063

Step-by-step explanation:

A random variable X has a Poisson distribution and it is referred to as Poisson random variable if and only if its probability distribution is given by

$p(x;\lambda)=\frac{\lambda e^{-\lambda}}{x!}$ for x = 0, 1, 2, ...

where $\lambda$ , the mean number of successes.

(a) To find the probability of exactly three flaws in 150 m of cable, we first need to find the mean number of flaws in 150 m, we know that the mean number of flaws in 50 m of cable is 1.2, so the mean number of flaws in 150 m of cable is $1.2 \cdot 3 =3.6$

The probability of exactly three flaws in 150 m of cable is

$P(X=3)=p(3;3.6)=\frac{3.6^3e^{-3.6}}{3!} \approx 0.21246$

(b) The probability of at least two flaws in 100m of cable is,

we know that the mean number of flaws in 50 m of cable is 1.2, so the mean number of flaws in 100 m of cable is $1.2 \cdot 2 =2.4$

$P(X\geq 2)=1-P(X$

$P(X\geq 2)=1-p(0;2.4)-p(1;2.4)\\\\P(X\geq 2)=1-\frac{2.4^0e^{-2.4}}{0!}-\frac{2.4^1e^{-2.4}}{1!}\\\\P(X\geq 2)\approx 0.69155$

(c) The probability of exactly one flaw in the first 50 m of cable, and exactly one flaw in the second 50 m of cable is

$P(X=1)=p(1;1.2)=\frac{1.2^1e^{-1.2}}{1!}\\P(X=1)\approx 0.36143$

The occurrence of flaws in the first and second 50 m of cable are independent events. Therefore the probability of exactly one flaw in the first 50 m and exactly one flaw in the second 50 m is

$(0.36143)(0.36143) = 0.13063$