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Romashka-Z-Leto [24]
3 years ago
12

Eileen randomly chooses 15 seeds from the bag of flower seeds she has left over from last year, and 10 seeds from the bag she bo

ught this year. Which sampling technique did she use to collect the 25 seeds in her sample?
A.simple random sampling
B.systematic sampling
C.convenience sampling
D.stratified random sampling
Mathematics
2 answers:
amm18123 years ago
5 0
I believe it is D, stratified random sampling.
Hope this helps!!

harina [27]3 years ago
5 0
The correct answer would be D.).
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In september, seattle's daily high temperature has a mean of 70 °f and a standard deviation of 7 °f. What are the mean, standard
Anna [14]
Mean= is the summation of the maximum temperatures ÷ the numbers of times the temperatures where recorded

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8 0
3 years ago
At a hockey game, a vendor sold a combined total of 148 sodas and hot dogs. The number of sodas sold was three times the number
AleksandrR [38]

Answer:

37 hot dogs and 111 sodas

Step-by-step explanation:

148 divided by 4= 37

148 - 37= 111

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3 0
2 years ago
Read 2 more answers
Plzz anyone solve all answers plzzzzzzzz​
algol13

You posted a lot of problems here. In the future please only post one problem at a time. Thank you.

I'll do the first two problems to get you started. Hopefully it will help you finish off the rest of the questions.

==========================================

Problem 1

{18, a, b, -3} is an arithmetic sequence or arithmetic progression (AP).

This means we have some number d added on to each term to get the next term.

first term = 18

second term = first term + d = 18+d = a

third term = second term + d = (18+d)+d = 18+2d = b

fourth term = third term + d = (18+2d)+d = 18+3d = -3

----

Let's solve that last equation for d

18+3d = -3

18+3d-18 = -3-18

3d = -21

3d/3 = -21/3

d = -7

----

The value d = -7 tells us to add -7 to each term to get the next term. In other words, we subtract 7 from each term to get the next term

first term = 18

second term = first term + d = 18+d = 18+(-7) = 18-7 = 11

third term = second term + d = 11+d = 11+(-7) = 11-7 = 4

fourth term = third term + d = 4+d = 4+(-7) = 4-7 = -3

----

We see that a = 11 and b = 4 are the second and third terms respectively.

Therefore, a+b = 11+4 = 15

-------------

<h3>Answer: 15</h3>

==========================================

Problem 2

A multiple of 4 is in the form 4*n for some integer n, ie n is a whole number.

We want to know which values of 4*n are between 10 and 250.

----

Divide both 10 and 250 by 4 to get the following

10/4 = 2.5

250/4 = 62.5

If n = 2, then 4*n = 4*2 = 8 is not between 10 and 250; however n = 3 will make 4*n = 4*3 = 12 to be between 10 and 250. We see that n = 3 is the smallest possible allowed value.

If n = 62, then 4*n = 4*62 = 248 is between 10 and 250; while n = 63 will make 4*n too big because 4*63 = 252. The largest n can get is n = 62

----

The question posed in question 2 is equivalent to asking the following: "How many values are in the set {3, 4, 5, ..., 60, 61, 62}?"

You could count all of the values in the set, but that exercise is very tedious busywork. There's a much faster way. First lets consider the set below

{a, a+1, a+2, ..., b-2, b-1, b}

where a,b are integers. Basically this set starts at 'a', counts up until we get to 'b'. The handy formula

c = b-a+1

will provide the exact count of values in the set {a, a+1, a+2, ..., b-2, b-1, b}

----

In this case, a = 3 and b = 62, making

c = b-a+1

c = 62-3+1

c = 60

There are 60 values in the set {3, 4, 5, ..., 60, 61, 62}

There are 60 multiples of four that are between 10 and 250.

-------------

<h3>Answer: 60</h3>
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The coefficient of the second term is the 3 in (4x+3y)
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