Let's see here. Sarah is one person. Her three friends are three more. That means 4 people are decorating picture frames. However, we only have 3 rolls of ribbon! We can give an entire roll to each person, but one person would have no roll to decorate with! What do we do? We have to divide this. By definition, divide means to separate something(in this case, ribbon rolls) into (equal) parts. So, we must divide them so everyone gets an EQUAL ribbon roll length.
The probability of drawing only one red card in two draws without replacement is given by:
![P(1\ red)=\frac{26C1\times26C1}{52C2}=0.51](https://tex.z-dn.net/?f=P%281%5C%20red%29%3D%5Cfrac%7B26C1%5Ctimes26C1%7D%7B52C2%7D%3D0.51)
The probability of drawing two red cards in two draws without replacement isgiven by:
![P(2\ red)=\frac{26C2\times26C0}{52C2}=0.245](https://tex.z-dn.net/?f=P%282%5C%20red%29%3D%5Cfrac%7B26C2%5Ctimes26C0%7D%7B52C2%7D%3D0.245)
The events 'draw one red card' and 'draw two red cards' are mutually exclusive. Therefore the probability of drawing at least one red card is 0.51 + 0.245 = 0.755.
If your talking about solar power its because its just not practical solar power doesnt put out nearly enough power for us to be able to run widespread as of yet new and more efficient solar power inc
Answer: 2500 years
Step-by-step explanation:
I'm not quite sure if I'm doing this right myself but I'll give it a shot.
We use this formula to find half-life but we can just plug in the numbers we know to find <em>t</em>.
![A(t)=A_{0}(1/2)^t^/^h](https://tex.z-dn.net/?f=A%28t%29%3DA_%7B0%7D%281%2F2%29%5Et%5E%2F%5Eh)
We know half-life is 5730 years and that the parchment has retained 74% of its Carbon-14. For
let's just assume that there are 100 original atoms of Carbon-14 and for A(t) let's assume there are 74 Carbon-14 atoms AFTER the amount of time has passed. That way, 74% of the C-14 still remains as 74/100 is 74%. Not quite sure how to explain it but I hope you get it. <em>h</em> is the last variable we need to know and it's just the half-life, which has been given to us already, 5730 years, so now we have this.
![74=100(1/2)^t^/^5^7^3^0](https://tex.z-dn.net/?f=74%3D100%281%2F2%29%5Et%5E%2F%5E5%5E7%5E3%5E0)
Now, solve. First, divide by 100.
![0.74=(0.5)^t^/^5^7^3^0](https://tex.z-dn.net/?f=0.74%3D%280.5%29%5Et%5E%2F%5E5%5E7%5E3%5E0)
Take the log of everything
![log(0.74)=\frac{t}{5730} log(0.5)](https://tex.z-dn.net/?f=log%280.74%29%3D%5Cfrac%7Bt%7D%7B5730%7D%20log%280.5%29)
Divide the entire equation by log (0.5) and multiply the entire equation by 5730 to isolate the <em>t</em> and get
![5730\frac{log(0.74)}{log(0.5)} =t](https://tex.z-dn.net/?f=5730%5Cfrac%7Blog%280.74%29%7D%7Blog%280.5%29%7D%20%3Dt)
Use your calculator to solve that giant mess for <em>t </em>and you'll get that <em>t</em> is roughly 2489.128182 years. Round that to the nearest hundred years, and you'll find the hopefully correct answer is 2500 years.
Really hope that all the equations that I wrote came out good and that that's right, this is definitely the longest answer I've ever written.
Answer:
A
Step-by-step explanation: