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Elden [556K]
3 years ago
5

,,sarah is building a birdhouse the nail she uses are one inch long.the wood board is 1 foot long . how many times smaller are t

he nails
Mathematics
2 answers:
Amiraneli [1.4K]3 years ago
4 0

Answer:

The board is 12 times larger

if u n\

tamaranim1 [39]3 years ago
3 0
The board is 12x larger in length than the nails. (Since the width of the board and nails are unknown, I am assuming you mean the length, not area difference)
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5(x – 6) + 3x – 2

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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
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