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3 years ago

What is 439 over 937 in simplest form

Mathematics

1 answer:

Licemer1 [7]3 years ago

4 0

I do believe it is 0.46852

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In the figure above, quadrilateral ABCD is a parallelogram. Let x represent the measure of angle GBF, y represent the measure of

ira [324]

Answers:
measure angle x = 40°
measure angle y = 35°
measure angle z = 55°

Explanation:
Part (a): getting angle x:
In triangle BED, we have:
measure angle BED = 90°
measure angle BDE = 50°
Therefore:
measure angle DBE = 180 - (90+50) = 40°
Now, we have angle DBE and angle GBF vertically opposite angles.
This means that they are both equal. Therefore angle GBF = 40°
Since angle GBF is x, therefore:
x = 40°

Part (b): getting angle y:
We know that the sum of measures of angles on a straight line is 180.
This means that:
angle GBF + angle GBC + angle CBE = 180
We have:
angle GBF = 40°
angle GBC = 105°
angle CBE = y
Therefore:
40 + 105 + y = 180
y = 35°

Part (c): getting angle z:
In triangle BCE, we have:
measure angle BCE = z
measure angle BEC = 90°
measure angle CBE = 35°
Therefore:
z + 90 + 35 = 180
z = 55°

Hope this helps :)

8 0

3 years ago

1/a+1/b=c solve for b how would i do this

olganol [36]

a cheap way to go about it is, let's do away with the denominators, and we'll do so by simply multiplying both sides by the the LCD of all fractions, in this case, that'd be "ab", so we multiply both sides by "ab".

$\bf \cfrac{1}{a}+\cfrac{1}{b}=c\implies\stackrel{\textit{multiplying both sides by }\stackrel{LCD}{ab}}{ab\left( \cfrac{1}{a}+\cfrac{1}{b} \right)=ab(c)}\implies b+a=abc \\\\\\ a=abc-b\implies a=\stackrel{\textit{common factor}}{b(ac-1)}\implies \cfrac{a}{ac-1}=b$

8 0

3 years ago

A man wins in a gambling game if he gets two heads in five flips of a biased coin. the probability of getting a head with the co

Semenov [28]

The probability the man will win will be 13.23%. And the probability of winning if he wins by getting at least four heads in five flips will be 36.01%.

<h3>How to find that a given condition can be modeled by binomial distribution?</h3>

Binomial distributions consist of n independent Bernoulli trials.

Bernoulli trials are those trials that end up randomly either on success (with probability p) or on failures( with probability 1- p = q (say))

P(X = x) = ⁿCₓ pˣ (1 - p)⁽ⁿ⁻ˣ⁾

A man wins in a gambling game if he gets two heads in five flips of a biased coin. the probability of getting a head with the coin is 0.7.

Then we have

p = 0.7

n = 5

Then the probability the man will win will be

P(X = 2) = ⁵C₂ (0.7)² (1 - 0.7)⁽⁵⁻²⁾

P(X = 2) = 10 x 0.49 x 0.027

P(X = 2) = 0.1323

P(X = 2) = 13.23%

Then the probability of winning if he wins by getting at least four heads in five flips will be

P(X = 4) = ⁵C₄ (0.7)⁴ (1 - 0.7)⁽⁵⁻⁴⁾

P(X = 4) = 5 x 0.2401 x 0.3

P(X = 4) = 0.3601

P(X = 4) = 36.01%

Learn more about binomial distribution here:

brainly.com/question/13609688

#SPJ1

4 0

2 years ago

10 POINTS!!!!<br><br> Name an angle that is vertical with ∠BOC.

Komok [63]

Vertical angle is the angle that is opposite to a specific angle

Thus the angle would be <FOE
or you can name it as <EOF
Makes no difference.

8 0

3 years ago

Evaluate the integral ∫2032x2+4dx. Your answer should be in the form kπ, where k is an integer. What is the value of k? (Hint: d

faltersainse [42]

Here is the correct computation of the question;

Evaluate the integral :

$\int\limits^2_0 \ \dfrac{32}{x^2 +4} \ dx$

Your answer should be in the form kπ, where k is an integer. What is the value of k?

(Hint: $\dfrac{d \ arc \ tan (x)}{dx} =\dfrac{1}{x^2 + 1}$ )

k = 4

(b) Now, lets evaluate the same integral using power series.

$f(x) = \dfrac{32}{x^2 +4}$

Then, integrate it from 0 to 2, and call it S. S should be an infinite series

What are the first few terms of S?

Answer:

(a) The value of k = 4

(b)

$a_0 = 16\\ \\ a_1 = -4 \\ \\ a_2 = \dfrac{12}{5} \\ \\a_3 = - \dfrac{12}{7} \\ \\ a_4 = \dfrac{12}{9}$

Step-by-step explanation:

(a)

$\int\limits^2_0 \dfrac{32}{x^2 + 4} \ dx$

$= 32 \int\limits^2_0 \dfrac{1}{x+4}\ dx$

$=32 (\dfrac{1}{2} \ arctan (\dfrac{x}{2}))^2__0$

$= 32 ( \dfrac{1}{2} arctan (\dfrac{2}{2})- \dfrac{1}{2} arctan (\dfrac{0}{2}))$

$= 32 ( \dfrac{1}{2}arctan (1) - \dfrac{1}{2} arctan (0))$

$= 32 ( \dfrac{1}{2}(\dfrac{\pi}{4})- \dfrac{1}{2}(0))$

$= 32 (\dfrac{\pi}{8}-0)$

$= 32 ( (\dfrac{\pi}{8}))$

$= 4 \pi$

The value of k = 4

(b) $\dfrac{32}{x^2+4}= 8 - \dfrac{3x^2}{2^1}+ \dfrac{3x^4}{2^3}- \dfrac{3x^6}{2x^5}+ \dfrac{3x^8}{2^7} -... \ \ \ \ \ (Taylor\ \ Series)$

$\int\limits^2_0 \dfrac{32}{x^2+4}= \int\limits^2_0 (8 - \dfrac{3x^2}{2^1}+ \dfrac{3x^4}{2^3}- \dfrac{3x^6}{2x^5}+ \dfrac{3x^8}{2^7} -...) dx$

$S = 8 \int\limits^2_0dx - \dfrac{3}{2^1} \int\limits^2_0 x^2 dx + \dfrac{3}{2^3}\int\limits^2_0 x^4 dx - \dfrac{3}{2^5}\int\limits^2_0 x^6 dx+ \dfrac{3}{2^7}\int\limits^2_0 x^8 dx-...$

$S = 8(x)^2_0 - \dfrac{3}{2^1*3}(x^3)^2_0 +\dfrac{3}{2^3*5}(x^5)^2_0- \dfrac{3}{2^5*7}(x^7)^2_0+ \dfrac{3}{2^7*9}(x^9)^2_0-...$

$S= 8(2-0)-\dfrac{1}{2^1}(2^3-0^3)+\dfrac{3}{2^3*5}(2^5-0^5)- \dfrac{3}{2^5*7}(2^7-0^7)+\dfrac{3}{2^7*9}(2^9-0^9)-...$

$S= 8(2-0)-\dfrac{1}{2^1}(2^3)+\dfrac{3}{2^3*5}(2^5)- \dfrac{3}{2^5*7}(2^7)+\dfrac{3}{2^7*9}(2^9)-...$

$S = 16-2^2+\dfrac{3}{5}(2^2) -\dfrac{3}{7}(2^2) + \dfrac{3}{9}(2^2) -...$

$S = 16-4 + \dfrac{12}{5}- \dfrac{12}{7}+ \dfrac{12}{9}-...$

$a_0 = 16\\ \\ a_1 = -4 \\ \\ a_2 = \dfrac{12}{5} \\ \\a_3 = - \dfrac{12}{7} \\ \\ a_4 = \dfrac{12}{9}$

6 0

3 years ago