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3 years ago

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The US postal service delivered 7.14 x 1010 pieces of mail in the month of December and 3.21 x 1010 in the month of January. Wha

t is the total mail delivered for these two months? A) 1.035 x 109 B) 1.035 x 1010 C) 1.035 x 10-10 D) 1.035 x 1011

2 answers:

nydimaria [60]3 years ago

8 0

The correct option is: D) $1.035 \times 10^1^1$

<em><u>Explanation</u></em>

Number of mails delivered in the month of December $=7.14 \times 10^1^0$

and number of mails delivered in the month of January $= 3.21\times 10^1^0$

For finding the total number of mails delivered for these two months, <u>we need to add $7.14 \times 10^1^0$ and $3.21\times 10^1^0$ </u>

$(7.14 \times 10^1^0)+(3.21\times 10^1^0)\\ \\ = (7.14+3.21)\times 10^1^0\\ \\ = 10.35\times 10^1^0\\ \\ =\frac{10.35}{10}\times 10^1^1 = 1.035 \times 10^1^1$

So, the total mail delivered for these two months is $1.035 \times 10^1^1$

Lelu [443]3 years ago

6 0

D) 1.035*10^11

Hope this helps

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Part 2: The probability of X≤2 or X≥4 is 0.5.

Part 3: The value of marginal probability of y is $f_y(y)=\frac{10-y}{32}$ for $2\leq y\leq 10$

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Part 9:The value of Cov(x,y) is 18.8886.

Part 10:X and Y are not independent variables as $f_{xy}(x,y)\neq f_x(x).f_y(y)\\$

Step-by-step explanation:

As the complete question is here, however some of the values are not readable, thus the question is found online and is attached herewith.

From the given data, the joint distribution is given as

$f(x,y)=\frac{1}{16}$ for $2\leq y\leq 2x\leq 10$

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$f_x(x)=\int\limits^{\infty}_{-\infty} {f(x,y)} \, dy$

Here the limits for y are $2\leq y\leq 2x$ So the equation becomes

$f_x(x)=\int\limits^{\infty}_{-\infty} {f(x,y)} \, dy\\f_x(x)=\int\limits^{2x}_{2} \frac{1}{16} \, dy\\f_x(x)=\frac{1}{16} (2x-2)\\f_x(x)=\frac{x-1}{8} \,\,\,\,\,\,\,\,\,\,\,\, for \,\,\,\,\,\,\,\,\,\ 1\leq x\leq 5$

Part 2:

The probability is given as

$P(X\leq 2 U X\geq 4)=\int\limits^2_1 {f_x(x)} \, dx +\int\limits^5_4 {f_x(x)} \, dx\\P(X\leq 2 U X\geq 4)=\int\limits^2_1 {\frac{x-1}{8}} \, dx +\int\limits^5_4 {\frac{x-1}{8}} \, dx\\P(X\leq 2 U X\geq 4)=\frac{1}{16}+\frac{7}{16}\\P(X\leq 2 U X\geq 4)=0.5$

So the probability of X≤2 or X≥4 is 0.5.

Part 3:

The distribution of y is given as

$f_y(y)=\int\limits^{\infty}_{-\infty} {f(x,y)} \, dx$

Here the limits for x are $y/2\leq x\leq 5$ So the equation becomes

$f_y(y)=\int\limits^{\infty}_{-\infty} {f(x,y)} \, dx\\f_y(y)=\int\limits^{5}_{y/2} \frac{1}{16} \, dx\\f_y(y)=\frac{1}{16} (5-\frac{y}{2})\\f_y(y)=\frac{10-y}{32} \,\,\,\,\,\,\,\,\,\,\,\, for \,\,\,\,\,\,\,\,\,\ 2\leq y\leq 10$

So the value of marginal probability of y is $f_y(y)=\frac{10-y}{32}$ for $2\leq y\leq 10$

Part 4

The value is given as

$E(y)=\int\limits^{10}_2 {yf_y(y)} \, dy\\E(y)=\int\limits^{10}_2 {y\frac{10-y}{32}} \, dy\\E(y)=\frac{1}{32}\int\limits^{10}_2 {10y-y^2} \, dy\\E(y)=4.6667$

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Part 5

This is given as

$f_{xy}(x)=\frac{f_{xy}(x,y)}{f_y(y)}\\f_{xy}(x)=\frac{\frac{1}{16}}{\frac{10-y}{32}}\\f_{xy}(x)=\frac{2}{10-y}$

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Part 6

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Part 7

The value is given as

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Part 8

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So the value of E(x,y) is 36

Part 9

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So the value of Cov(x,y) is 18.8886

Part 10

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$f_x(x).f_y(y)=\frac{x-1}{8}\frac{10-y}{32} \\$

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