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3 years ago

What is 20 fifths meter in centimeters?

2 answers:

8 0

First put the 20 fifths metres into something a little more recognisable:
20 x 1/5 metres
This is also 20/5 metres
And as 20/5 = 4, 20 fifths metres = 4 metres.

In each metre, there are 100 centimetres,
so 4 metres to 100 centimetres = 4 x 100 = 400 centimetres

So 20 fifths metres is 400 centimetres

Wittaler [7]3 years ago

3 0

* ^^ I was wrong, I read the question wrongly.

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Find the volume with radius 4 m and height 4 m .Use pencil and paper. How does the volume change when the height is tripled? Whe

VikaD [51]

Answer:

Step-by-step explanation:

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3 years ago

Suppose that the number of drivers who travel between a particular origin and destination during a designated time period has a

kipiarov [429]

Answer:

a) P(k≤11) = 0.021

b) P(k>23) = 0.213

c) P(11≤k≤23) = 0.777

P(11<k<23) = 0.699

d) P(15<k<25)=0.687

Step-by-step explanation:

a) What is the probability that the number of drivers will be at most 11?

We have to calculate P(k≤11)

$P(k\leq11)=\sum_0^{11} P(k$

$P(k=0) = 20^0e^{-20}/0!=1 \cdot 0.00000000206/1=0\\\\P(k=1) = 20^1e^{-20}/1!=20 \cdot 0.00000000206/1=0\\\\P(k=2) = 20^2e^{-20}/2!=400 \cdot 0.00000000206/2=0\\\\P(k=3) = 20^3e^{-20}/3!=8000 \cdot 0.00000000206/6=0\\\\P(k=4) = 20^4e^{-20}/4!=160000 \cdot 0.00000000206/24=0\\\\P(k=5) = 20^5e^{-20}/5!=3200000 \cdot 0.00000000206/120=0\\\\P(k=6) = 20^6e^{-20}/6!=64000000 \cdot 0.00000000206/720=0\\\\P(k=7) = 20^7e^{-20}/7!=1280000000 \cdot 0.00000000206/5040=0.001\\\\$

$P(k=8) = 20^8e^{-20}/8!=25600000000 \cdot 0.00000000206/40320=0.001\\\\P(k=9) = 20^9e^{-20}/9!=512000000000 \cdot 0.00000000206/362880=0.003\\\\P(k=10) = 20^{10}e^{-20}/10!=10240000000000 \cdot 0.00000000206/3628800=0.006\\\\P(k=11) = 20^{11}e^{-20}/11!=204800000000000 \cdot 0.00000000206/39916800=0.011\\\\$

$P(k\leq11)=\sum_0^{11} P(k$

b) What is the probability that the number of drivers will exceed 23?

We can write this as:

$P(k>23)=1-\sum_0^{23} P(k=x_i)=1-(P(k\leq11)+\sum_{12}^{23} P(k=x_i))$

$P(k=12) = 20^{12}e^{-20}/12!=8442485.238/479001600=0.018\\\\P(k=13) = 20^{13}e^{-20}/13!=168849704.75/6227020800=0.027\\\\P(k=14) = 20^{14}e^{-20}/14!=3376994095.003/87178291200=0.039\\\\P(k=15) = 20^{15}e^{-20}/15!=67539881900.067/1307674368000=0.052\\\\P(k=16) = 20^{16}e^{-20}/16!=1350797638001.33/20922789888000=0.065\\\\P(k=17) = 20^{17}e^{-20}/17!=27015952760026.7/355687428096000=0.076\\\\P(k=18) = 20^{18}e^{-20}/18!=540319055200533/6402373705728000=0.084\\\\$

$P(k=19) = 20^{19}e^{-20}/19!=10806381104010700/121645100408832000=0.089\\\\P(k=20) = 20^{20}e^{-20}/20!=216127622080213000/2432902008176640000=0.089\\\\P(k=21) = 20^{21}e^{-20}/21!=4322552441604270000/51090942171709400000=0.085\\\\P(k=22) = 20^{22}e^{-20}/22!=86451048832085300000/1.12400072777761E+21=0.077\\\\P(k=23) = 20^{23}e^{-20}/23!=1.72902097664171E+21/2.5852016738885E+22=0.067\\\\$

$P(k>23)=1-\sum_0^{23} P(k=x_i)=1-(P(k\leq11)+\sum_{12}^{23} P(k=x_i))\\\\P(k>23)=1-(0.021+0.766)=1-0.787=0.213$

c) What is the probability that the number of drivers will be between 11 and 23, inclusive? What is the probability that the number of drivers will be strictly between 11 and 23?

Between 11 and 23 inclusive:

$P(11\leq k\leq23)=P(x\leq23)-P(k\leq11)+P(k=11)\\\\P(11\leq k\leq23)=0.787-0.021+ 0.011=0.777$

Between 11 and 23 exclusive:

$P(11< k$

d) What is the probability that the number of drivers will be within 2 standard deviations of the mean value?

The standard deviation is

$\mu=\lambda =20\\\\\sigma=\sqrt{\lambda}=\sqrt{20}= 4.47$

Then, we have to calculate the probability of between 15 and 25 drivers approximately.

$P(15$

$P(k=16) = 20^{16}e^{-20}/16!=0.065\\\\P(k=17) = 20^{17}e^{-20}/17!=0.076\\\\P(k=18) = 20^{18}e^{-20}/18!=0.084\\\\P(k=19) = 20^{19}e^{-20}/19!=0.089\\\\P(k=20) = 20^{20}e^{-20}/20!=0.089\\\\P(k=21) = 20^{21}e^{-20}/21!=0.085\\\\P(k=22) = 20^{22}e^{-20}/22!=0.077\\\\P(k=23) = 20^{23}e^{-20}/23!=0.067\\\\P(k=24) = 20^{24}e^{-20}/24!=0.056\\\\$

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3 years ago

Add a T-shirt stand on the boardwalk five of the last 10 shirt shirt sold we read what is the experimental probability that the

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4 years ago

A store item is originally priced at $80. The item goes on sale at a 25% discount. Two weeks later that sales price is marked do

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3 years ago

Pamela deposits $4,000 in an account that earns interest at 5% APR compounded weekly. What is her ending balance after 2 years?

Triss [41]

The answer is $4419.76

A = P (1 + r/n)^nt

Therefore,

A = 4000 (1 + 0.05/52)^52 x 2
A = 4000 (1.00096)^104
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Hope that helps!
(I'll take brainliest thank you ;))

6 0

4 years ago

Read 2 more answers