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jenyasd209 [6]
3 years ago
15

1.3

Mathematics
1 answer:
krok68 [10]3 years ago
4 0

Answer:

The bottle that costs R59,99 for 375 g is cheaper

Step-by-step explanation:

The given costs of the bottles of honey are given as follows;

The cost of one of the bottles of honey = R89,99 for 530 g

The cost of the other bottle of honey = R59,99 for 357 g

Let 'x' represents the bottle of honey that costs R89,99 and let 'y' represent the bottle of honey that costs R59,99, we have;

The unit cost (cost per gram) for the bottle of honey, 'x', C/gₓ = R89,99/(530 g) ≈ R 0.16979/g

The unit cost (cost per gram) for the bottle of honey, 'y', C/g_y = R59,99/(357 g) ≈ R 0.168/g

Given that the unit cost of a bottle 'y' ≈ R 0.168/g is less than the unit cost of a bottle of 'x' ≈ R 016979/g, a bottle 'y' is cheaper than the bottle 'x'

Therefore;

The bottle that costs R59,99 for 375 g is cheaper than the bottle that costs R89,99 for 530 g.

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When two identical ions are separated by a distance of 4.2×10−10 , the electrostatic force each exerts on the other is 5.2×10−9
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Suppose that W1, W2, and W3 are independent uniform random variables with the following distributions: Wi ~ Uni(0,10*i). What is
nadya68 [22]

I'll leave the computation via R to you. The W_i are distributed uniformly on the intervals [0,10i], so that

f_{W_i}(w)=\begin{cases}\dfrac1{10i}&\text{for }0\le w\le10i\\\\0&\text{otherwise}\end{cases}

each with mean/expectation

E[W_i]=\displaystyle\int_{-\infty}^\infty wf_{W_i}(w)\,\mathrm dw=\int_0^{10i}\frac w{10i}\,\mathrm dw=5i

and variance

\mathrm{Var}[W_i]=E[(W_i-E[W_i])^2]=E[{W_i}^2]-E[W_i]^2

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E[W_1+W_2+W_3]=E[W_1]+E[W_2]+E[W_3]=5+10+15=30

and

\mathrm{Var}[W_1+W_2+W_3]=E\left[\big((W_1+W_2+W_3)-E[W_1+W_2+W_3]\big)^2\right]

\mathrm{Var}[W_1+W_2+W_3]=E[(W_1+W_2+W_3)^2]-E[W_1+W_2+W_3]^2

We have

(W_1+W_2+W_3)^2={W_1}^2+{W_2}^2+{W_3}^2+2(W_1W_2+W_1W_3+W_2W_3)

E[(W_1+W_2+W_3)^2]

=E[{W_1}^2]+E[{W_2}^2]+E[{W_3}^2]+2(E[W_1]E[W_2]+E[W_1]E[W_3]+E[W_2]E[W_3])

because W_i and W_j are independent when i\neq j, and so

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giving a variance of

\mathrm{Var}[W_1+W_2+W_3]=\dfrac{3050}3-30^2=\dfrac{350}3

and so the standard deviation is \sqrt{\dfrac{350}3}\approx\boxed{116.67}

# # #

A faster way, assuming you know the variance of a linear combination of independent random variables, is to compute

\mathrm{Var}[W_1+W_2+W_3]

=\mathrm{Var}[W_1]+\mathrm{Var}[W_2]+\mathrm{Var}[W_3]+2(\mathrm{Cov}[W_1,W_2]+\mathrm{Cov}[W_1,W_3]+\mathrm{Cov}[W_2,W_3])

and since the W_i are independent, each covariance is 0. Then

\mathrm{Var}[W_1+W_2+W_3]=\mathrm{Var}[W_1]+\mathrm{Var}[W_2]+\mathrm{Var}[W_3]

\mathrm{Var}[W_1+W_2+W_3]=\dfrac{25}3+\dfrac{100}3+75=\dfrac{350}3

and take the square root to get the standard deviation.

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