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mr_godi [17]
3 years ago
9

Kendra bought 4 boxes of doughnuts. There were 6 doughnuts in each box. She divided the doughnuts equally among 8 people. How ma

ny doughnuts did each person receive? doughnuts
Mathematics
2 answers:
uysha [10]3 years ago
5 0
Each person would receive 3 doughnuts because 4x6= 24 divide that by 8 and you get 3!
Akimi4 [234]3 years ago
3 0
Each person received 3 doughnuts, do 6 times 4 which is 24 and divide that among 8, its 3

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Find the fifth term in the arithmetic sequence that begins at 1.5 and has a common difference of 6.6.
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Sarah invested £12000 in a unit trust 5 years ago, the value of the unit trust increased by 7% per annum for each of the last 3
trasher [3.6K]

Answer:

The current price of the unit trust =  £13,831.72

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Since it increased 7% per annum in last three years and decreased by 3% per annum before that, it implies that the unit trust decreased by 3% per annum for the first 2 years and then increased by 7% per annum for the next 3 years in the total 5 year period

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This amount then grows by 7% for the next 3 years making it = 11,290.80*(1+0.07)^3 =  £13,831.7155 = £13,831.72 (Rounded to 2 decimals)

The current price of the unit trust =  £13,831.72 (Rounded to 2 decimals)

6 0
3 years ago
Find the mean, variance &a standard deviation of the binomial distribution with the given values of n and p.
MrMuchimi
A random variable following a binomial distribution over n trials with success probability p has PMF

f_X(x)=\dbinom nxp^x(1-p)^{n-x}

Because it's a proper probability distribution, you know that the sum of all the probabilities over the distribution's support must be 1, i.e.

\displaystyle\sum_xf_X(x)=\sum_{x=0}^n\binom nxp^x(1-p)^{n-x}=1

The mean is given by the expected value of the distribution,

\mathbb E(X)=\displaystyle\sum_xf_X(x)=\sum_{x=0}^nx\binom nxp^x(1-p)^{n-x}
\mathbb E(X)=\displaystyle\sum_{x=1}^nx\frac{n!}{x!(n-x)!}p^x(1-p)^{n-x}
\mathbb E(X)=\displaystyle\sum_{x=1}^n\frac{n!}{(x-1)!(n-x)!}p^x(1-p)^{n-x}
\mathbb E(X)=\displaystyle np\sum_{x=1}^n\frac{(n-1)!}{(x-1)!((n-1)-(x-1))!}p^{x-1}(1-p)^{(n-1)-(x-1)}
\mathbb E(X)=\displaystyle np\sum_{x=0}^n\frac{(n-1)!}{x!((n-1)-x)!}p^x(1-p)^{(n-1)-x}
\mathbb E(X)=\displaystyle np\sum_{x=0}^n\binom{n-1}xp^x(1-p)^{(n-1)-x}
\mathbb E(X)=\displaystyle np\sum_{x=0}^{n-1}\binom{n-1}xp^x(1-p)^{(n-1)-x}

The remaining sum has a summand which is the PMF of yet another binomial distribution with n-1 trials and the same success probability, so the sum is 1 and you're left with

\mathbb E(x)=np=126\times0.27=34.02

You can similarly derive the variance by computing \mathbb V(X)=\mathbb E(X^2)-\mathbb E(X)^2, but I'll leave that as an exercise for you. You would find that \mathbb V(X)=np(1-p), so the variance here would be

\mathbb V(X)=125\times0.27\times0.73=24.8346

The standard deviation is just the square root of the variance, which is

\sqrt{\mathbb V(X)}=\sqrt{24.3846}\approx4.9834
7 0
3 years ago
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