4. If the number of degrees in the measure of one angle of a triangle is represented by 4x and the measure of a second

Mathematics

1 answer:

Gnom [1K]3 years ago

4 0

Answer:

$third \: angle \: = 180 \degree - (4x + y) \degree$

Step-by-step explanation:

$4x \degree + y \degree + third \: angle \: = 180 \degree \\ \\ third \: angle \: = 180 \degree - (4x + y) \degree \\ \\$

You might be interested in

5. When I sine an angle like 60°, it gives me a value like 0.866025403. What does this number represent?

PolarNik [594]

Answer:

See below

Step-by-step explanation:

The number you described is the same as $sin(60^{\circ})=\frac{\sqrt{3}}{2}$ where the sine of an angle is the ratio between a right triangle's opposite side to the angle and the hypotenuse. So, in this case, if we had a right triangle with a height of $\sqrt{3}$ units and a hypotenuse of 2 units, the ratio between the two sides will result in the value you provided. This right triangle in particular would be a 30-60-90 triangle.

In the case of a unit circle, it’s the y-coordinate of the point where a 60° angle in the standard position intersects a unit circle and a right triangle is created from that.

5 0

3 years ago

All athletes at the Olympic Games (OG) are tested for performance-enhancing steroid drug use. The basic Anabolic Steroid Test (A

Semenov [28]

Answer:

a ) the probability of using steroids and having a negative test is 0.5%

b) The probability of testing positive is 6.4%

c) The probability of not using steroids, given that the test is negative is 99.47%

d) No, they are not statistically indepent.

e) The probability that the athlete will either use steroids or test positive is 6.9%

Step-by-step explanation:

Let A be the event that the test result is positive and B the event that the athlete uses Steroids. We are given the following

$P(A|B) = 90%, P(A|B^c) = 2%, P(B) = 5%$

From which we deduce that

$P(A^c|B) = 10%, P(B^c) = 95%$

a) We are asked for the probability $P(A^c\cap B)$ . REcall the conditional probability formula that, given two events C,D the conditional probability $P(C|D) = \frac{P(C\cap D)}{P(D)}$ . Then we have that

$P(A^c\cap B) = P(A^c|B)P(B) = 10\% \cdot 5\%=0.5\%$ .

b) We are asked for the probability P(A). We can use the fact that given two mutually exclusive events(that is, whose intersection is empty) A,B the probability P(C) of an event is given by $P(C) = P(C|A)P(A)+P(C|B)P(B)$ . Then

$P(A) = P(A|B)P(B)+P(A|B^c)P(B^c) = 90\%\cdot5\% + 2\% \cdot 95% = 6.4\%$

c) We are asked for the probability P(B^c|A^c). Recall that $P(A|B) = \frac{P(B|A)P(A)}{P(B)}$ . Then

$P(B^c|A^c) = \frac{P(A^c|B^c)P(B^c)}{P(A^c)}= \frac{P(A^c|B^c)P(B^c)}{1-P(A)}= \frac{(1-P(A|B^c))P(B^c)}{1-P(A)}=\frac{98\%\cdot 95\%}{1-6.4\%}= 99.47\%$

d) We say that two events A,B are statistically indepent if P(A|B) = P(A). Note that from point B the probability of testing negative is 1- 6.4% = 93.6%. Since 93.6% is different from 99.47% this means that testing positive and using steroids are not statistically independent.

e) We are asked for the probability $P(A\cup B)$ . We use the following