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3 years ago

Which statement best compares a line and a point

1 answer:

Volgvan3 years ago

6 0

A line has 0 endpoints and extends forever in 2 directions and a point has 1 endpoint and extends forever in 0 directions.

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GIVING BRAINLIEST TO THE PERSON THAT CAN EXPLAIN THIS THE BEST (:

scZoUnD [109]

Answer:

14w

Step-by-step explanation:

Width = w

Length is 6 time its width. so, multiply w by 6

length = 6*w = 6w

Perimeter of rectangle = 2* (length + width)

= 2*(6w + w)

= 2*7w

= 14w

4 0

1 year ago

Find the simple interest on $2,219 principal, deposited for 6 years at a rate of 1.91%

Aleks04 [339]

Answer:

$13,568.30

Step-by-step explanation:

I = Prt

I = 2,219(1.0191)(6) = 13,568.30

6 0

3 years ago

(9) power 3 +(3) power 2+(6) power 3+(15)power 2=?

Kisachek [45]

Hey! So, here's a tip. When writing exponents, an easier way is to write a^b, rather than a to the b power. Besides that, here is your answer!

So-------

9^3=729

3^2=9

6^3=216

15^2=225

Now that we have that figured out, we can add them together, wish is simple. 729 + 9 + 216 + 225= 1,179.

Therefore, your final answer will be 1,174.

If you have any questions on this, I'm happy to help you. :)

5 0

2 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

2 years ago

Find f^3(x) for f(x)=4/x^2

vaieri [72.5K]

Answer:

f³(x) = 64/x⁶

Step-by-step explanation:

f³(x) = (f(x))³ = (4/x²)³ = 4³/(x²)³

f³(x) = 64/x⁶

_____

The applicable rules of exponents are ...

(a/b)^c = (a^c)/(b^c)

(a^b)^c = a^(bc)

7 0

3 years ago