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2 years ago

HI PLS HELP TIMED WILL GIVE BRAINLIEST 22 POINTS!! :((((

Mathematics

1 answer:

vampirchik [111]2 years ago

6 0

Answer:

growth

decay

growth

decay

Step-by-step explanation:

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How to change base of logarithms: Best answer gets 27 points and brainliest answer

AlekseyPX

Answer:

1/3

Step-by-step explanation:

To change from one base to another, we use the formula

Logb x = Loga x/Loga b

log1/9 (3^(1/3) /3)

log3 ((3^(1/3) /3))

-------------------------

log3 (1/9)

Log a /b = log a - log b

and 1/9 = 3^-2

log3 ((3^(1/3) ) - log3 (3)

-------------------------

log3 (3^-2)

log a^b = blog a

1/3 log3 (3 ) - log3 (3)

-------------------------

-2log3 (3)

We know log3 (3) =1

1/3 (1) - 1

-------------------------

-2 (1)

1/3 - 1

-------------------------

-2

-2/3

------

-2

Copy dot flip

-2/3 * -1/2

1/3

6 0

3 years ago

A pen company averages 1.2 defective pens per carton produced (200 pens). The number of defects per carton is Poisson distribute

nlexa [21]

Answer:

a. P(x = 0 | λ = 1.2) = 0.301

b. P(x ≥ 8 | λ = 1.2) = 0.000

c. P(x > 5 | λ = 1.2) = 0.002

Step-by-step explanation:

If the number of defects per carton is Poisson distributed, with parameter 1.2 pens/carton, we can model the probability of k defects as:

$P(k)=\frac{\lambda^{k}e^{-\lambda}}{k!}= \frac{1.2^{k}\cdot e^{-1.2}}{k!}$

a. What is the probability of selecting a carton and finding no defective pens?

This happens for k=0, so the probability is:

$P(0)=\frac{1.2^{0}\cdot e^{-1.2}}{0!}=e^{-1.2}=0.301$

b. What is the probability of finding eight or more defective pens in a carton?

This can be calculated as one minus the probablity of having 7 or less defective pens.

$P(k\geq8)=1-P(k$

$P(0)=1.2^{0} \cdot e^{-1.2}/0!=1*0.3012/1=0.301\\\\P(1)=1.2^{1} \cdot e^{-1.2}/1!=1*0.3012/1=0.361\\\\P(2)=1.2^{2} \cdot e^{-1.2}/2!=1*0.3012/2=0.217\\\\P(3)=1.2^{3} \cdot e^{-1.2}/3!=2*0.3012/6=0.087\\\\P(4)=1.2^{4} \cdot e^{-1.2}/4!=2*0.3012/24=0.026\\\\P(5)=1.2^{5} \cdot e^{-1.2}/5!=2*0.3012/120=0.006\\\\P(6)=1.2^{6} \cdot e^{-1.2}/6!=3*0.3012/720=0.001\\\\P(7)=1.2^{7} \cdot e^{-1.2}/7!=4*0.3012/5040=0\\\\$

$P(k$

c. Suppose a purchaser of these pens will quit buying from the company if a carton contains more than five defective pens. What is the probability that a carton contains more than five defective pens?

We can calculate this as we did the previous question, but for k=5.

$P(k>5)=1-P(k\leq5)=1-\sum_{k=0}^5P(k)\\\\P(k>5)=1-(0.301+0.361+0.217+0.087+0.026+0.006)\\\\P(k>5)=1-0.998=0.002$