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3 years ago

What does scaled copies

Mathematics

2 answers:

ICE Princess25 [194]3 years ago

6 0

Its an image that is a scaled copy of the original and the shape is stretched..

melomori [17]3 years ago

5 0

An image is a scaled copy of the original if the shape is stretched in a way that does not distort it.

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A centipede is crawling 6 inches per minute. how long will it take the centipede to travel 1 yard?

Ede4ka [16]

6 minutes, because 1 yd is equivalent to 3 feet, which is equivalent to 36 inches. If a centipede crawls at 6 inches per minute, you do 6 x 6, which equals 36, and get 6 minutes!

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3 years ago

At what point is the function y=(x+3)/(x^(2)-7x+12) continuous

nika2105 [10]

If a function is defined as

$h(x)=\dfrac{f(x)}{g(x)}$

where both $f(x),g(x)$ are continuous functions, then $h(x)$ is also continuous where defined, i.e. where $g(x)\neq0$

So, in your case, this function is continous everywhere, except where

$x^2-7x+12=0$

To solve this equation, we can use the formula $x^2-sx+p=0$

It means that, if the leading terms is 1, then the x coefficient is the opposite of the sum of the roots, and the constant term is the product of the roots.

So, we're looking for two terms whose sum is 7, and whose product is 12. These numbers are easily found to be 3 and 4.

So, this function is continuous for every real number different than 3 or 4.

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2 years ago

7j + 5 - 8k when j = 0.5 and k = 0.25

tekilochka [14]

Answer:

I have no clue

Step-by-step explanation:

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2 years ago

Read 2 more answers

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:

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

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