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iVinArrow [24]
3 years ago
9

REAL ANSWERS PLEASE...

Mathematics
2 answers:
mart [117]3 years ago
8 0

Answer:

1) a) Check Below b) 9 2) Dependent b) Dependent  3)P(Pair \:of \:Pants \:or\:Blue\:Clothing)=\frac{10+4+8}{26}=\frac{22}{26}=\frac{11}{13}  =0.85

Step-by-step explanation:

1)

Let's match the possibilities:

Salmon, Fries  Salmon, Cooked Carrots    Salmon, Coleslaw

Trout, Fries     Trout, Cooked Carrots      Trout, Coleslaw

Halibut, Fries   Halibut, Cooked Carrots   Halibut, Coleslaw

b) 9 choices  

__3_ * __3___=9

2) If the Probability of "Winning" (A)  happens independently from the Probability of "Playing at Home" (B) and the result of P(A) does not influence P(B). We can calculate the probability of Winning and Playing at Home:

P(A\cap B) = P(A)\times P(B)

Rewriting the Chart

Home\\\\P(Win)=0.2\\P(Loss)=0.6\\P(Total)=0.8

Away:\\P(Win)=0.05\\P(Loss)=0.1\\P(Total)=0.15

Total\\\\P(Win)= 0.25\\P(Loss)=0.7\\P(Total)=1.00

a) Since the Total Events were not found by the product of "winning" and "playing at home" we cannot say these are independent events.

Dependent Event

b) Dependent Events

Let's remember the number of matches goes decreasing since the first one,  and that's what makes them dependent in probability, in other words, the denominator (sample space) decreases in the product.

3a) Blue Piece of Clothing. The Clothing is made up by shirt and pant

We have 12 shirts and 14 pair of pants, totalizing 26 clothing.

But the Blue ones are 8 shirts and 10 pants,

Then, let's add the  the amount of 4 black pants to the numerator and then add it, since we are including besides the blue clothing, the black pants over the the whole number of clothing:

P(Blue) =\frac{18}{26}=\frac{9}{13}

P(Pair \:of \:Pants \:or\:Blue\:Clothing)=\frac{10+4+8}{26}=\frac{22}{26}=\frac{11}{13}  =0.85

cestrela7 [59]3 years ago
3 0
1.) - Salmon                Trout        Halibut
       /       |           \        /   |  \         /   |  \
   Fries Carrots Slaw  F   C  S     F  C   S
-There is 9 different options, assuming that since its a combination problem and not a sequence problem, order doesn't matter.

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Typing errors in a text are either nonword errors (as when "the" is typed as "teh") or word errors that result in a real but inc
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a) X is binomial with n = 10 and p = 0.3

Y is binomial with n = 10 and p = 0.7

b) The mean number of errors caught is 7.

The mean number of errors missed is 3.

c) The standard deviation of the number of errors caught is 1.4491.

The standard deviation of the number of errors missed is 1.4491.

Step-by-step explanation:

For each typing error, there are only two possible outcomes. Either it is caught, or it is not. The probability of a typing error being caught is independent of other errors. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

Probability of exactly x sucesses on n repeated trials, with p probability.

The expected value of the binomial distribution is:

E(X) = np

The standard deviation of the binomial distribution is:

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10 word errors.

This means that n = 10

(a) If X is the number of word errors missed, what is the distribution of X ?

Human proofreaders catch 70 % of word errors. This means that they miss 30% of errors.

So for X, p = 0.3.

The answer is:

X is binomial with n = 10 and p = 0.3.

If Y is the number of word errors caught, what is the distribution of Y ?

Human proofreaders catch 70 % of word errors.

So for Y, p = 0.7.

The answer is:

Y is binomial with n = 10 and p = 0.7

(b) What is the mean number of errors caught?

E(Y) = np = 10*0.7 = 7

The mean number of errors caught is 7.

What is the mean number of errors missed?

E(X) = np = 10*0.3 = 3

The mean number of errors missed is 3.

(c) What is the standard deviation of the number of errors caught?

\sqrt{V(Y)} = \sqrt{np(1-p)} = \sqrt{10*0.7*0.3} = 1.4491

The standard deviation of the number of errors caught is 1.4491.

What is the standard deviation of the number of errors missed?

\sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{10*0.3*0.7} = 1.4491

The standard deviation of the number of errors missed is 1.4491.

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