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3 years ago

13

Given the quadrant of q in standard position and a trigonometric function value of q , find the exact value for the indicated fu

nction. II, sin q = 0.4; sec q

1 answer:

Verizon [17]3 years ago

8 0

9514 1404 393

Answer:

-(5/21)√21

Step-by-step explanation:

In the second quadrant, the cosine and secant are negative, so the value of interest is ...

sec(q) = -1/√(1 -0.4²) = -1/√0.84 = -√(25/21)

sec(q) = (-5√21)/21

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Twenty eight divided by fourteen is two.

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Four buses carrying 146 high school students arrive to Montreal. The buses carry, respectively, 32, 44, 28, and 42 students. One

Naily [24]

Answer:

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

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

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

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

8 0

4 years ago

A baker bakes 50 muffins. 1/5 of the muffins are chocolate chip. 1/2 of the muffins are blueberry. The rest are cinnamon. How ma

vitfil [10]

Answer:

15 muffins are cinnamon.

Step-by-step explanation:

Given that:

Number of muffins baked = 50 muffins

Chocolate chip muffins = 1/5 of 50 = $\frac{1}{5}*50$

Chocolate chip muffins = 10 muffins

Blueberry muffins = 1/2 of 50 = $\frac{1}{2}*50$

Blueberry muffins = 25

Cinnamon muffins = Muffins baked - chocolate chip muffins - blueberry muffins

Cinnamon muffins = 50 - 10 - 25

Cinnamon muffins = 50 - 35 = 15

Hence,

15 muffins are cinnamon.

5 0

3 years ago

Two fractions are given. For each one, write its decimal equivalent and determine if the decimal is terminating or non terminati

ziro4ka [17]

Answer:

9/20 yes, 4/15 not. See below

Step-by-step explanation:

Pick 9/20 and multiply numerator and denominator by 5:

9/20 = 45/100

We know that if we divide a number by 100 we need to move the coma as two places left, so:

9/20 = 45/100 = 0.45

And this is a terminal decimal as we know where it ends.

On the other hand if we pick 4/15 let try to divide it (here I will do it 'manually'):

4 |_ 15

we can divide 4 by 15, so we use 40 and begin with a comma

40 |_ 15

0.

15 enters 2 times in 40 with a rest of 10, so:

40 |_ 15

30 0.2

100

100 divided by 15 is 6 and we have 10 as rest again, and again and again...

40 |_ 15

30 0.266.....

100

100

....

So, we will have 0.266666666666666 infinitely. The decimal for 4/15 is non terminating and is 0.26666666666666666...

6 0

3 years ago