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2 years ago

Given circle o, what are the values of x and y?

Mathematics

1 answer:

GrogVix [38]2 years ago

7 0

Answer:

x = 38°

y = 90°

Step-by-step explanation:

x is an inscribed angle, so, 78/2=39

y is an inscribed angle of which connects two endpoints of the diameter, so y = 90°

Answered by GAUTHMATH

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Use the fundamental trigonometric identities to determine the simplified form of the expression

taurus [48]

Cot (B) = Cos(B) / Sin(B)

Cot B * Sin B = cos(B) / Sin(B) * Sin(B) = Cos(B)

Answer: Cos(B)

5 0

2 years ago

Rachel is bowling with her friends. her bowling ball has a radius of 4 inches. as she bowls she tracks the location of the finge

Goshia [24]

Question Continuation

The finger hole changes by 45 degrees.

Define a function, f, that gives the height of the finger hole above the ground (in inches) in terms of the angle of rotation (measured in radians) it has swept out from the 12 o'clock position.

Answer:

f(θ) = r(1 + cos(θ)) for 0 ≤ θ ≤ π/4

Step-by-step explanation:

Given

Let represent radius

r = 4 inches

Considering that she starts tracking the location when the finger hole is at the 12 o'clock; this means that the angle measurement at this point is 0°.

Let θ represent the angle

At 12 o'clock mark

θ = 0

When the finger hole changes by 45 degrees

θ = 45°

Convert 45° to radians

θ = 45° * π/180

θ = π/4

So, angle θ is such that θ∈[0, π/4]

This can be represented as

0 ≤ θ ≤ π/4

Calculating the measure of f(θ) in polar coordinates

When θ = 0, f(θ) = r (i.e. the current position of the bowl)

When θ = π/2, f(θ) = rcosθ

This is so because f(θ), being the function of the height is a measure of the radius* cos(θ)

Taking measurement of f(θ) from 0 to π/2

f(θ) = r + rcosθ

f(θ) = r(1 + cos(θ))

So, f(θ) = r(1 + cos(θ)) for 0 ≤ θ ≤ π/4

7 0

2 years ago

If a fair coin is flipped 15 times, what is the probability that there are more heads than tails?

ludmilkaskok [199]

Answer:

The probability that there are more heads than tails is equal to $\dfrac{1}{2}$ .

Step-by-step explanation:

Since the number of flips is an odd number, there can't be an equal number of heads and tails. In other words, there are either

more tails than heads, or,
more heads than tails.

Let the event that there are more heads than tails be $A$ . $\lnot A$ (i.e., not A) denotes that there are more tails than heads. Either one of these two cases must happen. As a result, $P(A) + P(\lnot A) = 1$ .

Additionally, since this coin is fair, the probability of getting a head is equal to the probability of getting a tail on each toss. That implies that (for example)

the probability of getting 7 heads out of 15 tosses will be the same as
the probability of getting 7 tails out of 15 tosses.

Due to this symmetry,

the probability of getting more heads than tails (A is true) is equal to
the probability of getting more tails than heads (A is not true.)

In other words $P(A) = P(\lnot A)$ .

Combining the two equations:

$\left\{\begin{aligned}&P(A) + P(\lnot A) = 1 \cr &P(A) = P(\lnot A)\end{aligned}\right.$ ,

$P(A) = P(\lnot A) = \dfrac{1}{2}$ .

In other words, the probability that there are more heads than tails is equal to $\dfrac{1}{2}$ .

This conclusion can be verified using the cumulative probability function for binomial distributions with $\dfrac{1}{2}$ as the probability of success.

$\begin{aligned}P(A) =& P(n \ge 8) \cr =& \sum \limits_{i = 8}^{15} {15 \choose i} (0.5)^{i} (0.5)^{15 - i}\cr =& \sum \limits_{i = 8}^{15} {15 \choose i} (0.5)^{15}\cr =& (0.5)^{15} \left({15 \choose 8} + {15 \choose 9} + \cdots + {15 \choose 15}\right) \cr =& (0.5)^{15} \left({15 \choose (15 - 8)} + {15 \choose (15 - 9)} + \cdots + {15 \choose (15 - 15)} \right) \cr =& (0.5)^{15} \left({15 \choose 7} + {15 \choose 6} + \cdots + {15 \choose 0}\right)\end{aligned}$