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densk [106]
3 years ago
12

The table represents the relationship between a length measured in meters and the same length measured in kilometers.

Mathematics
1 answer:
Bess [88]3 years ago
4 0

Answer:

Q1 1000     1

    3500    3.5

   500       .5

   75   .075

   1        .001

   x       1/1000x or .001x        Q2    Y=  1/1000x

Step-by-step explanation:

just search up 1000 meters to 1 kilometer and go from there :)  

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Multiply the following rational expressions and simplify the result
GarryVolchara [31]

Answer:

Step-by-step explanation:

We have to solve the given expression,

\frac{9y-33y^2-3y^4}{100-49y^2}.\frac{7y^2+17y+10}{14y^2+28y}

\frac{9y-33y^2-3y^4}{100-49y^2}.\frac{7y^2+17y+10}{14y^2+28y} = \frac{-y(-9+33y+3y^3)}{100-49y^2}.\frac{7y^2+17y+10}{14y(y+2)}

                                   = \frac{-y(-9+33y+3y^3)}{(10-7y)(10+7y)}.\frac{7y^2+10y+7y+10}{14y(y+2)}

                                   = \frac{-y(-9+33y+3y^3)}{(10-7y)(10+7y)}.\frac{y(7y+10)+1(7y+10)}{14y(y+2)}

                                   = \frac{-y(-9+33y+3y^3)}{(10-7y)(10+7y)}.\frac{(y+1)(7y+10)}{14y(y+2)}

                                   = \frac{-3y(-3+11y+y^3)}{(10-7y)}.\frac{(y+1)}{14y(y+2)}

                                   = \frac{-3(-3+11y+y^3)}{(10-7y)}.\frac{(y+1)}{14(y+2)}

                                   = \frac{3(3-11y-y^3)(y+1)}{(10-7y)(14(y+2)}

3 0
3 years ago
Why is not reasonable to say that 4.23 is less than 4.135
AysviL [449]

Simply because 4.135 is a larger number than 4.23.

8 0
3 years ago
What fraction of a pound is eighty pence
AnnZ [28]
4/5 is the the fraction of a pound is eighty pence
8 0
3 years ago
Read 2 more answers
There are 5 pens in a container on your desk. Among them, 3 will write well but 2 have defective ink cartridges. You will select
Norma-Jean [14]

Answer:

Kindly check explanation

Step-by-step explanation:

Total Number of pens = 5

Number of defective pens = 2

Number of non-defective pens = 3

A.) number of defective pens selected :

X : {0, 1, 2}

It is possible that no defective pen will be selected ; 1 defective will be chosen or both pens are defective.

2.)

Defective as Success (since selecting a defective pen is the point of interest.

3.)

Since selection is done without replacement

Probability of success per selection is different for each selection ;

Number of defective = 2

Number of observations = 5

P(success on first selection) = 2/5

P(success on second selection) = 1/4

Hence, X is not well approximated by a binomial random variable.

4.) if selection is done with replacement ; then then the probability of success per selection will be the Same for each selection made. Hence, X will be well approximated by a binomial Random variable.

5.) If sampling is done without replacement, then the the hypergeometric function will be a more effective approximation.

7 0
3 years ago
An article reports that 1 in 500 people carry the defective gene that causes inherited colon cancer. In a sample of 2000 individ
Tema [17]

Answer:

a) P=0.558

b) P=0.021

Step-by-step explanation:

We can model this random variable as a Poisson distribution with parameter λ=1/500*2000=4.

The approximate distribution of the number who carry this gene in a sample of 2000 individuals is:

P(x=k)=\frac{\lambda^ke^{-\lambda}}{k!} =\frac{4^ke^{-4}}{k!}

a) We can calculate that the approximate probability that between 4 and 9 (inclusive) as:

P(4\leq x\leq 9)=\sum_{k=4}^9P(k)\\\\\\ P(4)=4^{4} \cdot e^{-4}/4!=256*0.0183/24=0.195\\\\P(5)=4^{5} \cdot e^{-4}/5!=1024*0.0183/120=0.156\\\\P(6)=4^{6} \cdot e^{-4}/6!=4096*0.0183/720=0.104\\\\P(7)=4^{7} \cdot e^{-4}/7!=16384*0.0183/5040=0.06\\\\P(8)=4^{8} \cdot e^{-4}/8!=65536*0.0183/40320=0.03\\\\P(9)=4^{9} \cdot e^{-4}/9!=262144*0.0183/362880=0.013\\\\\\

P(4\leq x\leq 9)=\sum_{k=4}^9P(k)=0.195+0.156+0.104+0.060+0.030+0.013=0.558

b) The approximate probability that at least 9 carry the gene is:

P(x\geq9)=1-P(x\leq 8)\\\\\\

P(0)=4^{0} \cdot e^{-4}/0!=1*0.0183/1=0.018\\\\P(1)=4^{1} \cdot e^{-4}/1!=4*0.0183/1=0.073\\\\P(2)=4^{2} \cdot e^{-4}/2!=16*0.0183/2=0.147\\\\P(3)=4^{3} \cdot e^{-4}/3!=64*0.0183/6=0.195\\\\P(4)=4^{4} \cdot e^{-4}/4!=256*0.0183/24=0.195\\\\P(5)=4^{5} \cdot e^{-4}/5!=1024*0.0183/120=0.156\\\\P(6)=4^{6} \cdot e^{-4}/6!=4096*0.0183/720=0.104\\\\P(7)=4^{7} \cdot e^{-4}/7!=16384*0.0183/5040=0.06\\\\P(8)=4^{8} \cdot e^{-4}/8!=65536*0.0183/40320=0.03\\\\

P(x\geq9)=1-P(x\leq 8)\\\\P(x\geq9)=1-(0.018+0.073+0.147+0.195+0.195+0.156+0.104+0.060+0.030)\\\\P(x\geq9)=1-0.979=0.021

8 0
3 years ago
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