To do this you times 3 by 4 and add it on to the numerator of the fraction, to turn it into a top heavy fraction:
3*4 = 12
1+12/3 = 13/3
Then you multiply the numerator by 21 to work out what 21 times the fraction is:
13*21/3 = 273/3
Then you can divide 273 by 3 to get the final answer:
273/3 = 91
He will have 91 peaches overall.
Hope this helps! :)
Answer:
132 minutes
Step-by-step explanation:
1 Km = (60)(8)+15=495 seconds
4 mornings = (495)(4) =1980 seconds
4 weeks = (1980)(4) = 7920 sec.
7920/60 = 132 minutes
1). $2,700 is (2,700/6,000) = 45% of 6,000.
It took 6 years to earn it, so it earned (45%/6) = 7.5% each year.
That's the simple-interest rate.
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2). The investment earns 4.5% each year, so it earned 9% in 2 years.
The $1,150.65 is the 9% of the original investment. Call it V.
1150.65 = 0.09 x V .
Divide each side by 0.09 :
V = 1150.65 / 0.09 = $12,785 .
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3). There are 18 marbles in the bag all together.
a). 8 of them are green. If you close your eyes and pull out 1 marble,
the probability that it's a green one is
8/18 = 4/9 = 44.4% .
b). If it was a red marble, and you put it in your pocket, then
there are only 17 in the bag now, and 8 of them are still green.
If you pull another one, the probability that it's green is
8/17 = 47.1% .
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5). There are 10 marbles in the bag all together.
(The second sentence doesn't say anything, and doesn't mean anything.)
If you pull out a silver marble and put it in your pocket, then
there are 9 marbles in the bag, and 2 of them are orange.
The probability of pulling an orange marble now is
2/9 = 22.2% .
Answer:
a) 7.79%
b) 67.03%
c) Cumulative Distribution Function
Step-by-step explanation:
We are given the following in the question:
where x is the duration of a call, in minutes.
a) P( calls last between 2 and 3 minutes)
![=\displaystyle\int^3_2 p(x)~ dx\\\\= \displaystyle\int^3_20.1e^{-0.1x}~dx\\\\=\Big[-e^{-0.1x}\Big]^3_2\\\\=-\Big[e^{-0.3}-e^{-0.2}\Big]\\\\= 0.0779\\=7.79\%](https://tex.z-dn.net/?f=%3D%5Cdisplaystyle%5Cint%5E3_2%20p%28x%29~%20dx%5C%5C%5C%5C%3D%20%5Cdisplaystyle%5Cint%5E3_20.1e%5E%7B-0.1x%7D~dx%5C%5C%5C%5C%3D%5CBig%5B-e%5E%7B-0.1x%7D%5CBig%5D%5E3_2%5C%5C%5C%5C%3D-%5CBig%5Be%5E%7B-0.3%7D-e%5E%7B-0.2%7D%5CBig%5D%5C%5C%5C%5C%3D%200.0779%5C%5C%3D7.79%5C%25)
b) P(calls last 4 minutes or more)
![=\displaystyle\int^{\infty}_4 p(x)~ dx\\\\= \displaystyle\int^{\infty}_40.1e^{-0.1x}~dx\\\\=\Big[-e^{-0.1x}\Big]^{\infty}_4\\\\=-\Big[e^{\infty}-e^{-0.4}\Big]\\\\=-(0- 0.6703)\\= 0.6703\\=67.03\%](https://tex.z-dn.net/?f=%3D%5Cdisplaystyle%5Cint%5E%7B%5Cinfty%7D_4%20p%28x%29~%20dx%5C%5C%5C%5C%3D%20%5Cdisplaystyle%5Cint%5E%7B%5Cinfty%7D_40.1e%5E%7B-0.1x%7D~dx%5C%5C%5C%5C%3D%5CBig%5B-e%5E%7B-0.1x%7D%5CBig%5D%5E%7B%5Cinfty%7D_4%5C%5C%5C%5C%3D-%5CBig%5Be%5E%7B%5Cinfty%7D-e%5E%7B-0.4%7D%5CBig%5D%5C%5C%5C%5C%3D-%280-%090.6703%29%5C%5C%3D%200.6703%5C%5C%3D67.03%5C%25)
c) cumulative distribution function
