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3 years ago

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According to a survey by the park ranger, there is a 75% chance that hikers will see a deer on a park trail. Which simulation co

uld you use to answer questions about whether a hiker will see a deer?

1 answer:

Vika [28.1K]3 years ago

5 0

Answer:

Im bored wanna talk?

Step-by-step explanation:

You might be interested in

I need help with 1-4

seraphim [82]

Answer:

1) Point form (1,-4) Equation form x=1,y=-4

2) Point form (1,-2) Equation form x=1,y=-2

3) y=-23+7x

4) Point form (2,-2) Equation form x=2,y=-2

Step-by-step explanation:

For number 3 thats all i could get

3 0

3 years ago

Pls show full working out

sweet-ann [11.9K]

Answer:

Question 1a. i) $x=1.8m$

Question 1a. ii) $Volume=27.6m^3$

Question 1b) $Volume=65.9m^3$

Explanation:

<u><em>Question 1 a. i) Find the value of x.</em></u>

$tan(\theta )=\dfrac{opposite\text{ }leg}{adjacent\text{ }leg}$

For the smalll triangle you can write:

$tan(\theta )=\dfrac{x}{1m}$

For tthe big triangle:

$tan(\theta )=\dfrac{x+2.7m}{2.5m}$

Substitute:

$\dfrac{x}{1m}=\dfrac{x+2.7m}{2.5m}$

Solve for x:

$2.5x=x+2.7m\\\\2.5x-x=2.7m\\\\1.5x=2.7m\\\\x=2.7m/1.5\\\\x=1.8m$

<u><em>Question 1a ii) Find the volume of the frustrum</em></u>

Find the volume of a cone with height = 2.7m + 1.8m = 4.5m, and radius = 2.5m

Formula:

$Volume=(1/3)\pi \times radius^2\times height$

Substitute:

$Volume=(1/3)\pi \times (2.5m)^2\times 4.5m=9.375\pi m^3$

Find the volume of a cone with heigth = 1.8m and radius = 1m

$Volume=(1/3)\pi \times (1m)^2\times 1.8m=0.6\pi m^3$

Subtract the volume of the small cone from the volume of the big cone

$Volume\text{ }of\text{ }frustrum=9.375\pi m^3-0.6\pi m^3=8.775\pi m^3\approx 27.6m^3$

<u><em>Question 1b. Calculate the volume of the bin</em></u>

<u>i) Upper frustrum</u>

This is the same frustrum from the equation of above, thus ist volume is 27.6m³.

<u>ii) Lower frustrum</u>

$\dfrac{x}{2.0m}=\dfrac{x+2.4m}{2.5m}$

$2.5x=2(x+2.4m)\\\\2.5x=2x+4.8m\\\\0.5x=4.8m\\\\x=9.6m$

$Volume=(1/3)\pi \times (2.5m)^2\times (9.6m+2.4m)-(2.0m)^2\times (9.6m)$

$Volume=38.3m^3$

<u>iii) Add the volume of the two frustrums</u>

$Volume=27.6m^3+38.3m^3=65.9m^3$

6 0

3 years ago

A student performed the following steps to find the solution to the equation x^2 + 14x + 40 = 0. Where did the student go wrong?

Alja [10]

Answer:

A. in Step 3

Step-by-step explanation:

It is factored correctly. Then separated into two smaller equations correctly. But in solving x + 10 = 0, there is an error.

x + 10 = 0

Add -10 to both sides.

x = -10

So, the error is in step 3.

3 0

2 years ago

Read 2 more answers

The handbook of chemistry and physics lists the density of a certain liquid to be 0.7988 g/mL. Taylor experimentally finds this

yan [13]

Answer:

Yes

Step-by-step explanation:

0.7988-0.5= 0.7488

0.7988+0.5= 1.2988

0.7925 lies between these values.

however, are you certain that the question has 0.500 as the uncertainty not 0.050

6 0

2 years ago

Read 2 more answers

The variable x varies directly as the cube of y, and y varies directly as the square root of z. If x equals 1 when z equals 4, w

Aloiza [94]

Answer:

<h2> $z=36$ </h2>

Step-by-step explanation:

According to the question,

$x$ ∝ $y^3$ .......(1)

$y$ ∝ $\sqrt{z}$ .......(2)

From equation 1,2 let constant of proportionality be $k1,k2$ respectively.

⇒ $x=k1(y^3)$ .......(3)

⇒ $y=k2(\sqrt{z} )$ .......(4)

From the above equations putting 4 into 3,

$x=k1((k2\sqrt{z})^3) =k1.k2^3.(\sqrt{z})^3$

Let the new constant to the above equation be $k3$ ,

$x=k3(\sqrt{z})^3$

Given,if x=1, when z=4

$1=k3(\sqrt{4} )^3=k3(8)$

⇒ $k3=\frac{1}{8}$

Now if x=27, then z=?

$27=\frac{1}{8} (\sqrt{z} )^3$

⇒ $(\sqrt{z} )^3=27(8)$

⇒ $\sqrt{z}=3(2)=6$

$z=36$

4 0

3 years ago