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2 years ago

Let W be an acute angle. Use a calculator to approximate the measure of W to the nearest tenth of a degree.

Mathematics

1 answer:

ikadub [295]2 years ago

7 0

Answer: $57.3^{\circ}$

Step-by-step explanation:

$\cos W=0.54\\W=\cos^{-1}(0.54) \approx \boxed{57.3^{\circ}}$

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Answer the 5 questions Listed below Please Show your Work

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3 years ago

Use FOIL to explain how to find the product of (a + b)(a − b). Then describe a shortcut that you could use to get this product w

andriy [413]

Greetings!

FOIL stands for:

Front
Outside
Inside
Last

This tells which terms to multiply when using the Distributive Property.
(NOTE: Only applicable with 2-term Polynomials)

For Example:
$(a+b)(a-b)$

Multiply the Fronts of both Equations:
$(a*a)$

Multiply the Outsides of both Equations:
$(a*a)+(a*-b)$

Multiply the Insides of both Equations:
$(a*a)+(a*-b)+(b*a)$

Multiply the Lasts of both Equations:
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Simplify.
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$=a^2-b^2$

Alternative Method (My Prefered Method)
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Use Regular Distributive Property.
$a(a+b)-b(a+b)$

$a*a+a*b-b*a+-b*b$

Simplify.
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$a^2-b^2$

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3 years ago

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2 years ago

Let X equal the number of hours that a randomly selected person from City A reads the news online each day. Suppose that X has a

Nina [5.8K]

Answer:

$P(Y>X) = \frac{17}{32}$

Step-by-step explanation:

Recall that since X is uniformly distributed over the set [1,4] we have that the pdf of X is given by $f(x) = \frac{1}{3}$ if $1\leq x \leq 4$ and 0 otherwise. In the same manner, the pdf of Y is given by $g(y) = \frac{1}{4}$ if $1\leq y\leq 5$ and 0 otherwise.

Note that if Y is in the interval (4,5] then Y>X by default. So, in this case we have that P(Y>X| y in (4,5]) = 1. We want to calculate the probability of having Y in that interval . That is

$P(Y>4) = \int_{4}^{5}\frac{1}{4}dy = \frac{1}{4}$ . Thus, $P(Y\leq 4 ) = 1 - P(Y>4)= \frac{3}{4}$ .

We want to proceed as follows. Using the total probability theorem, given two events A, B we have that

$P(A) = P(A|B)\cdot P(B) + P(A|B^c) \cdot P(B^c)$ In this case, A is the event that Y>X and B is the event that Y is in the interval (4,5].

If we assume that X and Y are independent, then we have that the joint pdf of X,Y is given by $h(x,y) = f(x)g(y) = \frac{1}{12}$ when $1\leq x \leq 4, 1\leq y \leq 4$ . We can draw the region were Y>X and the function h(x,y) is different from 0. (The drawing is attached). This region is described as follows: $1\leq x \leq 4$ and $x\leq y \leq 4$ , then (the specifics of the calculations of the integrals are ommitted)

$P(Y>X| y \in [1,4)) = \int_{1}^{4}\int_{x}^{4}\frac{1}{12}dy dx = \frac{1}{12}\int_{1}^{4} (4-x) dx = \frac{1}{12}\left.(4x-\frac{x}{2})\right|_{1}^{4}= \frac{1}{12}(4\cdot 4 - \frac{4^2}{2}-(4-\frac{1}{2}) = \frac{9}{2\cdot 12} = \frac{3}{8}$

Thus,

$P(Y>X) = 1\cdot \frac{1}{4} + \frac{3}{8}\cdot \frac{3}{4} = \frac{17}{32}$

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4 years ago