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3 years ago

A teacher is writing an exam with 60

Mathematics

1 answer:

Mumz [18]3 years ago

7 0

20 is the answer your welcome you will get it right don’t worry

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A company offers courses to prepare students for the Graduate Management Admission Test (GMAT) has the following information abo

ki77a [65]

Answer:

OK aaaaaaaa. What is it

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3 years ago

A particular concentration of a chemical found in polluted water has been found to be lethal to 20% of the fish that are exposed

torisob [31]

Answer:

a) $P(X=14)=(20C14)(0.8)^{14} (1-0.8)^{20-14}=0.109$

b) $P(X\geq 10) = 1-P(X \leq 9) = 0.9994$

c) $P(X \leq 16)= 1-P(X>16) =1-P(X \geq 17)= 1- [P(X=17) +...+P(X=20)]=0.589$

d) $E(X)= np = 20 *0.8 = 16$

$Var(X) = np(1-p) = 20*0.8*(1-0.8) = 3.2$

Step-by-step explanation:

Previous concepts

The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".

Let X the random variable of interest, on this case we now that:

$X \sim Binom(n=20, p=1-0.2=0.8)$

The probability mass function for the Binomial distribution is given as:

$P(X)=(nCx)(p)^x (1-p)^{n-x}$

Where (nCx) means combinatory and it's given by this formula:

$nCx=\frac{n!}{(n-x)! x!}$

Part a

For this case we want this probability:

$P(X=14)$

And using the mass function we have this:

$P(X=14)=(20C14)(0.8)^{14} (1-0.8)^{20-14}=0.109$

Part b

For this case we want this probability:

$P(X \geq 10)$

And we can find this using the complement rule:

$P(X\geq 10) = 1-P(X$

$P(X=0)=(20C0)(0.8)^{0} (1-0.8)^{20-0}=1.05x10^{-14}$

$P(X=1)=(20C1)(0.8)^{1} (1-0.8)^{20-1}=8.39x10^{-13}$

$P(X=2)=(20C2)(0.8)^{2} (1-0.8)^{20-2}=3.19x10^{-11}$

$P(X=3)=(20C3)(0.8)^{3} (1-0.8)^{20-3}=7.65x10^{-10}$

$P(X=4)=(20C4)(0.8)^{4} (1-0.8)^{20-4}=1.30x10^{-8}$

$P(X=5)=(20C5)(0.8)^{5} (1-0.8)^{20-5}=1.66x10^{-7}$

$P(X=6)=(20C6)(0.8)^{6} (1-0.8)^{20-6}=1.66x10^{-6}$

$P(X=7)=(20C7)(0.8)^{7} (1-0.8)^{20-7}=1.33x10^{-5}$

$P(X=8)=(20C8)(0.8)^{8} (1-0.8)^{20-8}=8.65x10^{-5}$

$P(X=9)=(20C9)(0.8)^{9} (1-0.8)^{20-9}=0.00046$

And if we replace we got:

$P(X\geq 10) = 1-P(X \leq 9) = 0.9994$

Part c

For this case we want this probability:

$P(X \leq 16)$

And we can use the complement rule like this:

$P(X \leq 16)= 1-P(X>16) =1-P(X \geq 17)= 1- [P(X=17) +...+P(X=20)]$

$P(X=17)=(20C17)(0.8)^{17} (1-0.8)^{20-17}=0.205$

$P(X=18)=(20C18)(0.8)^{18} (1-0.8)^{20-18}=0.137$

$P(X=19)=(20C19)(0.8)^{19} (1-0.8)^{20-19}=0.0576$

$P(X=20)=(20C20)(0.8)^{20} (1-0.8)^{20-20}=0.0115$

And if we replace we got:

$P(X \leq 16)= 1-P(X>16) =1-P(X \geq 17)= 1- [P(X=17) +...+P(X=20)]=0.589$

Part d

The expected value is given by:

$E(X)= np = 20 *0.8 = 16$

$Var(X) = np(1-p) = 20*0.8*(1-0.8) = 3.2$

3 0

3 years ago

A company has an annual electric bill of $495,000. Of that amount, 3/5 was billed

disa [49]

Answer:

$74,250

Step-by-step explanation:

495,000x0.6=297,000

297,000/4=74,250

7 0

3 years ago

How do I find this help please

KIM [24]

Area of the sector = 1/2 * r^2 * theta where theta is the angle subtended by the arc at the center ( in radians)

In this case m < theta = 360 - 235 = 125 degrees or 2.182 radians

So, the area of shaded area = 1/2 * 20^2 * 2.182 = 436.4 in^2

8 0

4 years ago

Write 5.24 as a mixed number in simplest form. 5.24 =

NeTakaya

Step-by-step explanation:

5 +0.24 = 5 + 24 = 12 = 6

100 50 25

= 5 6

3 0

3 years ago