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3 years ago

A building lot in a city is shaped as a 30°-60°-90° triangle, like the figure shown.

Mathematics

2 answers:

Julli [10]3 years ago

4 0

Check the picture below.

In-s [12.5K]3 years ago

4 0

A) You can find the length of the side of the lot opposite the 60° angle (which we will call "x") as below:

Tan(α)=Opposite/Adjacent

α=60°
Opposite=41 feet
Adjacent=x

When you substitute these values into Tan(α)=Opposite/Adjacent, you obtain:

Tan(60°)=x/41
(41)Tan(60°)=x
x=41√3
x=71.01 feet

B) Now, you can find the length of the hypotenuse:

Cos(α)=Adjacent/Hypotenuse

α=60°
Adjacent=41 feet
Hypotenuse=y

When you substitute these values into Cos(α)=Adjacent/Hypotenuse, you obtain:

Cos(60°)=41/y
yCos(60°)=41
y=41/Cos(60°)
y=82 feet

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Out of 2,000 random but normally distributed numbers with a mean of 45 and a standard deviation of x, approximately 1,360 number

Lilit [14]

Let $Y$ denote the random variable representing a given number in the total set of numbers. We're told that $\dfrac{1360}{2000}=0.68$ of the numbers fall within a given range, so we know

$\mathbb P(40\le Y\le50)=0.68$

where $Y$ is normally distributed with mean 45 and an unknown variance $x^2$ .

Let's make the transformation to a random variable with a standard normal distribution:

$\mathbb P\left(\dfrac{40-45}x\le\dfrac{Y-45}x\le\dfrac{50-45}x\right)=\mathbb P\left(-\dfrac5x\le Z\le\dfrac5x\right)$

Since $Z$ is symmetric, we have

$\mathbb P(-\dfrac5x\le Z\le\dfrac5x\right)=2\mathbb P\left(0\le Z\le\dfrac5x\right)=2\bigg(\mathbb P\left(Z\le\dfrac5x\right)-\mathbb P(Z\le0)\bigg)$

The mean of $Z$ is 0, and by symmetry we know that exactly half of the distribution falls to the left of $Z=0$ , so $\mathbb P(Z\le0)=0.5$ . We're left with

$0.68=2\mathbb P\left(Z\le\dfrac5x\right)-2\cdot0.5$
$\implies\mathbb P\left(Z\le\dfrac5x\right)=0.84$

This probability corresponds to a value of $Z\approx0.9945$ , which means

$\dfrac5x\approx0.9945\implies x\approx5.0277$

7 0

3 years ago

Solve: <img src="https://tex.z-dn.net/?f=%5Cunderset%7Bx%5Crightarrow~3%7D%7B%5Clim%7D~%5Cdfrac%7B2x%5E2-18%7D%7Bx%5E2-3x

vodomira [7]

Hello, please consider the following.

$\displaystyle \lim_{x\rightarrow3}~\dfrac{2x^2-18}{x^2-3x} \\ \\ \\ =\lim_{x\rightarrow3}~\dfrac{2(x^2-3^2)}{x(x-3)} \\ \\ \\ =\lim_{x\rightarrow3}~\dfrac{2(x-3)(x+3)}{x(x-3)} \\ \\ \\ =\lim_{x\rightarrow3}~\dfrac{2(x+3)}{x} \\ \\ \\=\dfrac{2(3+3)}{3}\\ \\ \\=\dfrac{2*3*2}{3} =\Large \boxed{\sf \bf \ 4 \ }$

Thank you

8 0

3 years ago

I need help with these.

sattari [20]

Answer:

See below.

Step-by-step explanation:

5^2 + 12^2 = x^2

25 + 144 = x^2

169 = x^2

x = 13

3^2 + x^2 = [sqrt(10)]^2

9 + x^2 = 10

x^2 = 1

x = 1

1^2 + x^2 = 4^2

1 + x^2 = 16

x^2 = 15

x = sqrt(15)

[sqrt(27)]^2 + x^2 = 6^2

27 + x^2 = 36

x^2 = 9

x = 3

15/5 = c/sqrt(29)

5c = 15 * sqrt(29)

c = 3sqrt(29)

26/13 = x/12

2 = x/12

x = 24

7 0

3 years ago

Mr.kiniston jogged how many miles

Dimas [21]

The answer is letter G.

4 0

4 years ago

Read 2 more answers

PLZZZ HELP I'LL GIVE U A MEDAL N FAN U!!!

Goshia [24]

3.14 * 18=56.52

so your answer is c

hope i helped
good luck on your homework

4 0

3 years ago

Read 2 more answers