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valentina_108 [34]
3 years ago
11

Eric, Cam, and Brooke are playing a game of mini golf on a computer. The computer program for the game stores

Mathematics
1 answer:
andrew-mc [135]3 years ago
7 0

Answer:

Eric

Step-by-step explanation:

First, we determine the distance of each player to the hole using the distance formula.

Given points (x_1,y_1)$ and (x_2,y_2)

Distance=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}

<u>Eric</u>

The hole is at (3,2)

Eric's golf ball is at point (-6,6)

Distance=\sqrt{(-6-3)^2+(6-2)^2}

=\sqrt{(-9)^2+(4)^2}\\=\sqrt{97} \\\approx 9.85

<u>Cam</u>

The hole is at (3,2)

Cam's golf ball is at point (8,-6)

Distance=\sqrt{(8-3)^2+(-6-2)^2}

=\sqrt{(5)^2+(-8)^2}\\=\sqrt{89} \approx9.43

<u>Brooke</u>

The hole is at (3,2)

Brooke's golf ball is at point (10,8)

Distance=\sqrt{(10-3)^2+(8-2)^2}\\=\sqrt{(7)^2+(6)^2}\\=\sqrt{85} \approx 9.22

Since Eric's ball is the farthest from the hole, Eric should shoot next.

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Gennadij [26K]

Answer:

84

Step-by-step explanation:

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3 years ago
The length of a rectangle is 4 less than twice the width. If the perimeter of the rectangle is 130, find the length of the recta
Alchen [17]

Answer:

Let L be the length of the rectangle and w be the width of the rectangle.

"The lenght of a rectangle is one less than twice the width" means L=2W-1

Using perimeter formula of a rectangle which is P=2(L+W) you have:

P=2(L+W)

130=2(2W-1+W)

Solve equation above to find W

130=2(3W-1)

130=6W-2

130+2=6W

132=6W

W=22

From here you find L=2W-1=2x22-1=43

Only thing left is to find area A=LxW

4 0
3 years ago
Suppose we roll a fair die and let X represent the number on the die. (a) Find the moment generating function of X. (b) Use the
Likurg_2 [28]

Answer:

(a)  moment generating function for X is \frac{1}{6}\left(e^{t}+e^{2 t}+e^{2 t}+e^{4 t}+e^{5 t}+e^{6 t}\right)

(b) \mathrm{E}(\mathrm{X})=\frac{21}{6} \text { and } E\left(X^{2}\right)=\frac{91}{6}

Step-by step explanation:

Given X represents the number on die.

The possible outcomes of X are 1, 2, 3, 4, 5, 6.

For a fair die, P(X)=\frac{1}{6}

(a) Moment generating function can be written as M_{x}(t).

M_x(t)=\sum_{x=1}^{6} P(X=x)

M_{x}(t)=\frac{1}{6} e^{t}+\frac{1}{6} e^{2 t}+\frac{1}{6} e^{3 t}+\frac{1}{6} e^{4 t}+\frac{1}{6} e^{5 t}+\frac{1}{6} e^{6 t}

M_x(t)=\frac{1}{6}\left(e^{t}+e^{2 t}+e^{3 t}+e^{4 t}+e^{5 t}+e^{6 t}\right)

(b) Now, find E(X) \text { and } E\((X^{2}) using moment generating function

M^{\prime}(t)=\frac{1}{6}\left(e^{t}+2 e^{2 t}+3 e^{3 t}+4 e^{4 t}+5 e^{5 t}+6 e^{6 t}\right)

M^{\prime}(0)=E(X)=\frac{1}{6}(1+2+3+4+5+6)  

\Rightarrow E(X)=\frac{21}{6}

M^{\prime \prime}(t)=\frac{1}{6}\left(e^{t}+4 e^{2 t}+9 e^{3 t}+16 e^{4 t}+25 e^{5 t}+36 e^{6 t}\right)

M^{\prime \prime}(0)=E(X)=\frac{1}{6}(1+4+9+16+25+36)

\Rightarrow E\left(X^{2}\right)=\frac{91}{6}  

Hence, (a) moment generating function for X is \frac{1}{6}\left(e^{t}+e^{2 t}+e^{3 t}+e^{4 t}+e^{5 t}+e^{6 t}\right).

(b) \mathrm{E}(\mathrm{X})=\frac{21}{6} \text { and } E\left(X^{2}\right)=\frac{91}{6}

6 0
4 years ago
Please help me out please
Elanso [62]

Answer:

400 units³

Step-by-step explanation:

The volume (V) of the square pyramid is

V = \frac{1}{3} area of base × height (h)

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Consider the right triangle formed by a segment from the vertex to the midpoint of the base and the slant height ( the hypotenuse )

Using Pythagoras' identity on the right triangle

h² + 5² = 13²

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h² = 144 ( take the square root of both sides )

h = \sqrt{144} = 12

Area of square base = 10² = 100, hence

V = \frac{1}{3} × 100 × 12 = 4 × 100 = 400

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