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3 years ago

Find the value of x in each case.

Mathematics

1 answer:

Kisachek [45]3 years ago

6 0

Answer:

Step-by-step explanation:

As shown in the picture, you have:

CF and CD are perpendicular

=> angle BCF = 90 deg

=> angle BCE + angle ECF = angle BCF = 90 deg

or angle BCE + 64 = 90

=> angle BCE = 90 - 64 = 26 deg

Point A, B, and C are collinear

=> (angle ABE) + angle CBE = 180 deg (1)

Otherwise, as the property of the sum of 3 angles in a triangle:

(angle BEC + angle BCE) + angle CBE = 180 deg (2)

From (1) and (2) you would have:

angle ABE = angle BEC + angle BCE

3x = 26 + x

<=> 3x - x = 26

<=> 2x = 26

<=> x = 26/2

<=> x = 13

=> angle BEC = x = 13 deg

=> angle ABE = 3x = 3(13) = 39 deg

Hope this helps!

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The expected value of X is $E(X)=\frac{2754}{73} \approx 37.73$ and the variance of X is $Var(X)=\frac{226192}{5329} \approx 42.45$

The expected value of Y is $E(Y)=\frac{73}{2} \approx 36.5$ and the variance of Y is $Var(Y)=\frac{179}{4} \approx 44.75$

Step-by-step explanation:

(a) Let X be a discrete random variable with set of possible values D and probability mass function p(x). The expected value, denoted by E(X) or $\mu_x$ , is

$E(X)=\sum_{x\in D} x\cdot p(x)$

The probability mass function $p_{X}(x)$ of X is given by

$p_{X}(28)=\frac{28}{146} \\\\p_{X}(32)=\frac{32}{146} \\\\p_{X}(42)=\frac{42}{146} \\\\p_{X}(44)=\frac{44}{146}$

Since the bus driver is equally likely to drive any of the 4 buses, the probability mass function $p_{Y}(x)$ of Y is given by

$p_{Y}(28)=p_{Y}(32)=p_{Y}(42)=p_{Y}(44)=\frac{1}{4}$

The expected value of X is

$E(X)=\sum_{x\in [28,32,42,44]} x\cdot p_{X}(x)$

$E(X)=28\cdot \frac{28}{146}+32\cdot \frac{32}{146} +42\cdot \frac{42}{146} +44 \cdot \frac{44}{146}\\\\E(X)=\frac{392}{73}+\frac{512}{73}+\frac{882}{73}+\frac{968}{73}\\\\E(X)=\frac{2754}{73} \approx 37.73$

The expected value of Y is

$E(Y)=\sum_{x\in [28,32,42,44]} x\cdot p_{Y}(x)$

$E(Y)=28\cdot \frac{1}{4}+32\cdot \frac{1}{4} +42\cdot \frac{1}{4} +44 \cdot \frac{1}{4}\\\\E(Y)=146\cdot \frac{1}{4}\\\\E(Y)=\frac{73}{2} \approx 36.5$

(b) Let X have probability mass function p(x) and expected value E(X). Then the variance of X, denoted by V(X), is

$V(X)=\sum_{x\in D} (x-\mu)^2\cdot p(x)=E(X^2)-[E(X)]^2$

The variance of X is

$E(X^2)=\sum_{x\in [28,32,42,44]} x^2\cdot p_{X}(x)$

$E(X^2)=28^2\cdot \frac{28}{146}+32^2\cdot \frac{32}{146} +42^2\cdot \frac{42}{146} +44^2 \cdot \frac{44}{146}\\\\E(X^2)=\frac{10976}{73}+\frac{16384}{73}+\frac{37044}{73}+\frac{42592}{73}\\\\E(X^2)=\frac{106996}{73}$

$Var(X)=E(X^2)-(E(X))^2\\\\Var(X)=\frac{106996}{73}-(\frac{2754}{73})^2\\\\Var(X)=\frac{106996}{73}-\frac{7584516}{5329}\\\\Var(X)=\frac{7810708}{5329}-\frac{7584516}{5329}\\\\Var(X)=\frac{226192}{5329} \approx 42.45$

The variance of Y is

$E(Y^2)=\sum_{x\in [28,32,42,44]} x^2\cdot p_{Y}(x)$

$E(Y^2)=28^2\cdot \frac{1}{4}+32^2\cdot \frac{1}{4} +42^2\cdot \frac{1}{4} +44^2 \cdot \frac{1}{4}\\\\E(Y^2)=196+256+441+484\\\\E(Y^2)=1377$

$Var(Y)=E(Y^2)-(E(Y))^2\\\\Var(Y)=1377-(\frac{73}{2})^2\\\\Var(Y)=1377-\frac{5329}{4}\\\\Var(Y)=\frac{179}{4} \approx 44.75$

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