Ask question

Ask question

All categories

chubhunter [2.5K]

2 years ago

8

A camp counselor is trying to determine, of the places they have previously camped, where his campers' favorite camping spots ar

e. what is the population of the survey?
A. campers who have gone camping before

B. all campers

C. campers who have not gone camping before

D. all counselors

2 answers:

GarryVolchara [31]2 years ago

8 0

A

B is incorrect, because some of the campers may have never gone camping before, so they would be unable to say which is their favorite camping spot
C is incorrect because all of those campers have never gone camping before, so they would be unable to say which is their favorite camping spot
D is incorrect because she is asking the campers not the counselors

stira [4]2 years ago

3 0

Answer:

A

Step-by-step explanation:

Because how are they supposed to know their favorite camping spot if they've never been camping.

You might be interested in

How much water do you add to make 12% sugar syrup using 5g sugar?

Alchen [17]

so the syrup is just water and sugar, we know that the sugar percent is just 12%, and that happens to be 5grams.

now 100% - 12% is 88%, so the 88% leftover will just be water.

if 12% is 5g, how much is 88%?

$\bf \begin{array}{ccll} amount&\%\\ \cline{1-2} 5&12\\ x&88 \end{array}\implies \cfrac{5}{x}=\cfrac{12}{88}\implies \cfrac{5}{x}=\cfrac{3}{22}\implies 110=3x \\\\\\ \cfrac{110}{3}=x\implies 36\frac{2}{3}=x$

8 0

2 years ago

Read 2 more answers

A police officer, in her car, is watching a person in a sports car. Assume the locations of the cars are mapped onto the coordin

Paha777 [63]

is there and answers like mutipol choice

5 0

2 years ago

Read 2 more answers

Old faithful in Yellowstone National Park shoots water 60 feet into the air that casts a shadow of 42 feet. What is the height o

Serhud [2]

Answer:

The height of near by tree will be 90 feet.

Step-by-step explanation:

Considering the water triangle as shown in figure a.

As the old faithful shoots water 60 feet into the air that casts a shadow of 42 feet.

Considering the tree triangle as shown in figure a. The tree triangle has a shadow 63 feet long.

Let x be the height of near by tree that cost a shadow 63 feet long.

Assuming that the triangles are similar

So,

the ratios of similar triangles will be related as:

$\frac{63}{42}=\frac{x}{60}$

$\mathrm{Switch\:sides}$

$\frac{x}{60}=\frac{63}{42}$

${x}=60.\frac{63}{42}$

$x=90$ feet

Therefore, the height of near by tree will be 90 feet.

Keywords: ratio, similar triangle

Learn more about similar triangle from brainly.com/question/10161974

#learnwithBrainly

7 0

2 years ago

A cement bridge post is 24 inches square and 15 feet 9 inches in length. If the cement weighs 145 pounds per cubic foot, how muc

lisov135 [29]

Answer:

The answer is option (C), One cement bridge post weighs 9,135 pounds

Step-by-step explanation:

Step 1: Get the total volume of a cement bridge post

The cement bridge post is in the shape of a cuboid, therefor the volume of the cement bridge post can be expressed as;

Volume of cement bridge post=Base area×Length

where;

Base area=(24^2) inches square

1 foot=12 inches

Convert 24 inches to foot=24/12=2^2=4 feet²

Length=15 feet and 9 inches

1 foot=12 inches

Convert 9 inches to foot=9/12=0.75 feet

Total length=(15+0.75)=15.75 feet

replacing;

Volume of cement bridge post=(4×15.75)=63 cubic feet

Volume of cement bridge post=63 cubic feet

Step 2: Get the total weight of the cement bridge post

Total weight of the cement bridge post=Weight per cubic foot×total volume of the cement bridge post

where;

Weight per cubic foot=145 pounds per cubic foot

Total volume of the cement bridge post=63 cubic feet

replacing;

Total weight of the cement bridge post=(145×63)=9,135 pounds

One cement bridge post weighs 9,135 pounds

8 0

2 years ago

Find the equation of the plane through the point (2,5,7) that is parallel to the line r=(3i+2j−2k)+t(i+2j+9k) and perpendicular

insens350 [35]

The plane we want to find has general equation

$a(x-2)+b(y-5)+c(z-7)=0$

with $a,b,c$ not equal to 0, and has normal vector

$\vec n=a\,\vec\imath+b\,\vec\jmath+c\,\vec k$

$\vec n$ is perpendicular to both the normal vector of the other plane, which is $4\,\vec\imath+5\,\vec\jmath+6\,\vec k$ , as well as the tangent vector to the line $\vec r(t)$ , which is $\vec\imath+2\,\vec\jmath+9\,\vec k$ .

This means the dot product of $\vec n$ with either vector is 0, giving us

$\begin{cases}4a+5b+6c=0\\a+2b+9c=0\end{cases}$

Suppose we fix $c=1$ . Then the system reduces to

$\begin{cases}4a+5b=-6\\a+2b=-9\end{cases}$

and we get

$(4a+5b)-4(a+2b)=-6-4(-9)\implies-3b=30\implies b=-10$

$a+2(-10)=-9\implies a=11$

Then one equation for the plane could be

$\boxed{11(x-2)-10(y-5)+(z-7)=0}$

or in standard form,

$\boxed{11x-10y+z=-21}$

The solution is unique up to non-zero scalar multiplication, which is to say that any equation $(11x-10y+z)k=-21k$ would be a valid answer. For example, suppose we instead let $c=2$ ; then we would have found $a=22$ and $b=-20$ , but clearly dividing both sides of the equation

$22(x-2)-20(y-5)+2(z-7)=0$

by 2 gives the same equation as before.

7 0

2 years ago