How do I solve this?

A golfer hits an errant tee shot that lands in the rough. A marker in the center of the fairway is 150 yards from the center of

Gwar [14]

$\bold{\huge{\underline{ Solution}}}$

<h3>Given :-</h3>

A marker in the center of the fairway is 150 yards away from the centre of the green
While standing on the marker and facing the green, the golfer turns 100° towards his ball
Then he peces off 30 yards to his ball

We have to find the distance between the golf ball and the center of the green .

<h3>Let's Begin :-</h3>

Let assume that the distance between the golf ball and central of green is x

Here, 

Distance between marker and centre of green is 150 yards
That is, Height = 150 yards
For facing the green , The golfer turns 100° towards his ball
That is, Angle = 100°
The golfer peces off 30 yards to his ball
That is, Base = 30 yards

According to the law of cosine :-

$\bold{\red{ a^{2} = b^{2} + c^{2} - 2ABcos}}{\bold{\red{\theta}}}$

Here, a = perpendicular height
b = base
c = hypotenuse
cos theta = Angle of cosine

So, For Hypotenuse law of cosine will be :-

$\sf{ c^{2} = a^{2} + b^{2} - 2ABcos}{\sf{\theta}}$

Subsitute the required values,

$\sf{ x^{2} = (150)^{2} + (30)^{2} - 2(150)(30)cos}{\sf{100°}}$

$\sf{ x^{2} = 22500 + 900 - 900cos}{\sf{\times{\dfrac{5π}{9}}}}$

$\sf{ x^{2} = 22500 + 900 - 900( - 0.174)}$

$\sf{ x^{2} = 22500 + 900 + 156.6}$

$\sf{ x^{2} = 23556.6}$

$\bold{ x = 153.48\: yards }$

Hence, The distance between the ball and the center of green is 153.48 or 153.5 yards

5 0

2 years ago

Assume that women's heights are normally distributed with a mean given by mu equals 62.5 in,and a standard deviation given by

Misha Larkins [42]

Answer:

(a) 0.5899

(b) 0.9166

Step-by-step explanation:

Let X be the random variable that represents the height of a woman. Then, X is normally distributed with

$\mu$ = 62.5 in

$\sigma$ = 2.2 in

the normal probability density function is given by

$f(x) = \frac{1}{\sqrt{2\pi}2.2}\exp{-\frac{(x-62.5)^{2}}{2(2.2)^{2}}}$ , then

(a) $P(X < 63) = \int\limits_{-\infty}^{63}f(x) dx$ = 0.5899

(in the R statistical programming language) pnorm(63, mean = 62.5, sd = 2.2)

(b) We are seeking $P(\bar{X} < 63)$ where n = 37. $\bar{X}$ is normally distributed with mean 62.5 in and standard deviation $2.2/\sqrt{37}$ . So, the probability density function is given by

$g(x) = \frac{1}{\sqrt{2\pi}\frac{2.2}{\sqrt{37}}}\exp{-\frac{(x-62.5)^{2}}{2(2.2/\sqrt{37})^{2}}}$ , and

$P(\bar{X} < 63) = \int\limits_{-\infty}^{63}g(x)dx$ = 0.9166

(in the R statistical programming language) pnorm(63, mean = 62.5, sd = 2.2/sqrt(37))

You can use a table from a book to find the probabilities or a programming language like the R statistical programming language.

4 0

3 years ago

Please help me on #37. I know for A you would use the quadratic equation but I got the wrong answer.

ozzi

Δ = (8i)^2 - 4*(-25) => Δ = -36 +100 => Δ = 64 => x1 =(-8i + 8)/2 => x1 = -4i +4;
x2 = (-8i - 8)/2 => x2 = -4i-4; in this case, your solutions are complex conjugates.

6 0

3 years ago

What is the following sum in simplest form sqrt8+3sqrt2+sqrt32 please HURRY

iragen [17]

Answer:

9sqrt2.

Step-by-step explanation:

sqrt8 + 3sqrt2 + sqrt32

= sqrt4 * sqrt2 + 3 sqrt2 + sqrt16 * sqrt2

= 2sqrt2 + 3sqrt2 + 4 sqrt2

= 9 sqrt2.

3 0

3 years ago

Read 2 more answers

The length of a rectangle is four times its width. If the perimeter of the rectangle is 20 feet, how many feet long is the recta

Gemiola [76]

Largeur = x
longueur = 4 x

2 ( x + 4 x) = 20
2 x + 8 x = 20
10 x =20
x = 2

largeur = 2 et longueur = 8

8 0

3 years ago