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katen-ka-za [31]
3 years ago
13

Esmerelda is a math student who needs to gather a sample of 10 participants to conduct a survey about U.S. residents. She asks h

er 3 siblings, her 2 parents, and telephones 5 friends and asks them to complete her questionnaire. This is an example of a ______________________.
A. representative sample

B. convenience sample

C. self-selecting sample

D. systematic sample
Mathematics
1 answer:
Nuetrik [128]3 years ago
8 0

A convenience sample is made up of people who are easy to reach so it would be B

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a regular polygon has a perimeter of 36 inches. each side is exactly 6 inches long. which figure does this discribe?
fgiga [73]
Since a side is only 6 inches long and the total perimeter is 36; take 36/6 and you get 6. The only figure with six sides is a hexagon, thus your answer is hexagon.
3 0
3 years ago
There are 96 dogs at the dog park.
BARSIC [14]

Answer:

4 dogs :]

Step-by-step explanation:

1/8 brown dogs = 12 dogs

5/6 black dogs = 80 dogs

12 + 80 = 92

96 - 92 = 4

Therefore, 4 dogs are mixed colors.

Hope this helps, sorry if this is wrong! Have a great day! <)

6 0
3 years ago
Read 2 more answers
A. The actual building height will be 596 ft tall. You want your model to be 3 ft tall. What will the scale factor be? For this
Reil [10]

Answer:

mulltiply

Step-by-step explanation: you mulitply all

7 0
3 years ago
How many different linear arrangements are there of the letters a, b,c, d, e for which: (a a is last in line? (b a is before d?
inna [77]
A) Since a is last in line, we can disregard a, and concentrate on the remaining letters.
Let's start by drawing out a representation:

_ _ _ _ a

Since the other letters don't matter, then the number of ways simply becomes 4! = 24 ways

b) Since a is before d, we need to account for all of the possible cases.

Case 1: a d _ _ _ 
Case 2: a _ d _ _
Case 3: a _ _ d _
Case 4: a _ _ _ d

Let's start with case 1.
Since there are four different arrangements they can make, we also need to account for the remaining 4 letters.
\text{Case 1: } 4 \cdot 4!

Now, for case 2:
Let's group the three terms together. They can appear in: 3 spaces.
\text{Case 2: } 3 \cdot 4!

Case 3:
Exactly, the same process. Account for how many times this can happen, and multiply by 4!, since there are no specifics for the remaining letters.
\text{Case 3: } 2 \cdot 4!

\text{Case 4: } 1 \cdot 4!

\text{Total arrangements}: 4 \cdot 4! + 3 \cdot 4! + 2 \cdot 4! + 1 \cdot 4! = 240

c) Let's start by dealing with the restrictions.
By visually representing it, then we can see some obvious patterns.

a b c _ _

We know that this isn't the only arrangement that they can make.
From the previous question, we know that they can also sit in these positions:

_ a b c _
_ _ a b c

So, we have three possible arrangements. Now, we can say:
a c b _ _ or c a b _ _
and they are together.

In fact, they can swap in 3! ways. Thus, we need to account for these extra 3! and 2! (since the d and e can swap as well).

\text{Total arrangements: } 3 \cdot 3! \cdot 2! = 36
7 0
3 years ago
How do you solve this system: y= -3/2x, and 3x+2y=-4 by substitution?
Soloha48 [4]
Download Photomath it will answer it with step by step work
4 0
3 years ago
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