Plz help me!!!!!!!!!!!!!!!!!!!!

Kim rode her bike 135 miles in 9 weeks riding the same distance each week. Eric rode his bike 102 miles in 6 weeks the same dist

pentagon [3]

Kim = 135/9 = 15 miles per week
Eric = 102/6 = 17 miles per week

Eric rode more.

5 0

3 years ago

6 clients are throwing a party. Each cake serves 24 servings and has a party of 70 +3staff how many cakes are needed?

Airida [17]

4 cakes are needed fjfnccnfn

8 0

3 years ago

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Police records in the town of Saratoga show that 17 percent of the drivers stopped for speeding have invalid licenses. If 24 dri

Ymorist [56]

Answer:

Step-by-step explanation:

Given

17 % of the drivers stopped have invalid licenses

n=24 drivers are stopped

using Binomial distribution

$^nC_r(p)^r(q)^{n-r}$

here p=0.17

q=1-0.17=0.83

(a)none will have valid license i.e. r=0

$P(r=0)^{24}C_0(0.17)^0(0.83)^{24}=(0.83)^{24}=0.0114$

(b)Exactly one have invalid license i.e. r=1

$P(r=1)=^{24}C_1(0.17)^1(0.83)^{23}=24\times 0.17\times (0.83)^{23}$

$=0.0561$

(c)At least 2 will have invalid license i.e. $r\geq 2$

$P(r\geq 2)=1-P(r=0)-P(r=1)$

$P(r\geq 2)=1-0.0114-0.0561$

$P(r\geq 2)=0.932$

3 0

3 years ago

Scientist can determine the age of ancient objects by a method called radiocarbon dating. The bombardment of the upper atmospher

Artemon [7]

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 t.

$A(t)=A_{0}(1/2)^t^/^h$

We know half-life is 5730 years and that the parchment has retained 74% of its Carbon-14. For $A_{0$ 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. h 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.

$74=100(1/2)^t^/^5^7^3^0$

Now, solve. First, divide by 100.

$0.74=(0.5)^t^/^5^7^3^0$

Take the log of everything

$log(0.74)=\frac{t}{5730} log(0.5)$

Divide the entire equation by log (0.5) and multiply the entire equation by 5730 to isolate the t and get

$5730\frac{log(0.74)}{log(0.5)} =t$

Use your calculator to solve that giant mess for t and you'll get that t 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.

5 0

3 years ago

Greatest common factor of 24,40,56

Tema [17]

8 is the greatest common factor

8 0

3 years ago

Read 2 more answers