mark must leave it for 5.5 months or 5 and half moths to gain 5600 in interest .
<u>Step-by-step explanation:</u>
Here we have , mark invests 8000 in an account that pays 12% interest and 2000 in one that pays 8%. if he leaves the money in the accounts for the same length of time, We need to find how long must he leave it to gain 5600 in interest . Let's find out:
Let mark invests 8000 in an account that pays 12% interest and 2000 in one that pays 8% for time x months , So total interest gain is 5600 i.e.
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Therefore , mark must leave it for 5.5 months or 5 and half moths to gain 5600 in interest .
2:3:5.....added = 10
2/10(60) = 120/10 = 12 <=
3/10(60) = 180/10 = 18 <=
5/10(60) = 300/10 = 30 <=
Answer:
1/25x
Step-by-step explanation:
Answer:
30ft
Step-by-step explanation:
50- 20 = 30
hope it helps!
Answer:
So then the correct answer would be:
B) .9996
Step-by-step explanation:
The exact way to solve this problem is using the binomial distribution, assuming that our random variable of interest is "number of students living in apartments" represented by X and 
And we want this probability:
![P(200 \leq X \leq 400) [tex]In order to find this probability we can use the foloowing excel code:"=BINOM.DIST(400,600,0.4,TRUE)-BINOM.DIST(199,600,0.4,TRUE)"And we got:[tex] P(200 \leq X \leq 400) =0.999675[tex]But for this case the problem says that we need to approximate, so then we can use the normal approximation to the normal distribution. We need to check the conditions in order to use the normal approximation. [tex]np=600*0.4=240\geq 10](https://tex.z-dn.net/?f=%20P%28200%20%5Cleq%20X%20%5Cleq%20400%29%20%5Btex%5D%3C%2Fp%3E%3Cp%3EIn%20order%20to%20find%20this%20probability%20we%20can%20use%20the%20foloowing%20excel%20code%3A%3C%2Fp%3E%3Cp%3E%22%3DBINOM.DIST%28400%2C600%2C0.4%2CTRUE%29-BINOM.DIST%28199%2C600%2C0.4%2CTRUE%29%22%3C%2Fp%3E%3Cp%3EAnd%20we%20got%3A%3C%2Fp%3E%3Cp%3E%5Btex%5D%20P%28200%20%5Cleq%20X%20%5Cleq%20400%29%20%3D0.999675%5Btex%5D%3C%2Fp%3E%3Cp%3EBut%20for%20this%20case%20the%20problem%20says%20that%20we%20need%20to%20approximate%2C%20so%20then%20we%20can%20use%20the%20normal%20approximation%20to%20the%20normal%20distribution.%20%3C%2Fp%3E%3Cp%3EWe%20need%20to%20check%20the%20conditions%20in%20order%20to%20use%20the%20normal%20approximation.%0A%3C%2Fp%3E%3Cp%3E%5Btex%5Dnp%3D600%2A0.4%3D240%5Cgeq%2010)
So we see that we satisfy the conditions and then we can apply the approximation.
If we appply the approximation the new mean and standard deviation are:
And then 
And we are interested on the following probability:
So then the correct answer would be:
B) .9996
