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Firlakuza [10]
3 years ago
15

Which sampling would be appropriate to survey 120,000 people?

Mathematics
1 answer:
kipiarov [429]3 years ago
8 0

Answer: Id say b

Step-by-step explanation: a would be too high and a bit too detailed. Whereas c and d will be inaccurate.

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F(x) = 2x – 4/?<br> What are the domain and range
makkiz [27]

Answer:

domain

Step-by-step explanation:

range

8 0
3 years ago
What is the measure of angle TSU?​
goldfiish [28.3K]

Answer:

80

Step-by-step explanation:

Angle VSU = 100. Angles TSU is a linear pair with angle VSU so adding them together equals 180. 180-100=80.

7 0
3 years ago
Can someone please help me with this problem? You need to find the value of x and y by using the strategy of ELIMINATION. Please
Setler [38]

Answer:

x = 136/35; y = -⁹/₁₀

Step-by-step explanation:

(1)         7x + 8y = 20

(2)        7x – 2y = 29       Subtract (2) from (1)

                  10y =  -9        Divide each side by 10

(3)                 y =  -⁹/₁₀     Substitute (3) into (1)

      7x - 2(-⁹/₁₀) = 29

      7x + 18/10  = 29                       Subtract 18/10 from each side

      7x              =      29 - 18/10

      7x              = (290 - 18)/10

      7x              =          272/10        Divide each side by 7

        x              =           272/70

        x              =           136/35

x = 136/35; y = -⁹/₁₀

Check:

(1) 7(136/35) + 8(-⁹/₁₀) = 20

      136/5    -   72/10 = 20

      136/5    -   36/5  = 20

                      100/5  = 20

                       20      = 20

(2) 7(136/35) – 2(-⁹/₁₀)  = 29

       136/5    +    18/10 = 29

       136/5    +      ⁹/₅    = 29

                        145/5  = 29

                        29       = 29

5 0
3 years ago
Need help with geometry
ra1l [238]
The answer is x=3 because abt+tbc=abc
3 0
3 years ago
In 1981, the Australian humpback whale population was 350 and has increased at a rate of 14% each year since then.)
dimulka [17.4K]

Answer:

In 1981, the Australian humpback whale population was 350

Po = Initial population = 350

rate of increase = 14% annually

P(t) = Po*(1.14)^t

P(t) = 350*(1.14)^t

Where

t = number of years that have passed since 1981

Year 2000

2000 - 1981 = 19 years

P(19) = 350*(1.14)^19

P(19) = 350*12.055

P(19) = 4219.49

P(19) ≈ 4219

Year 2018

2018 - 1981 = 37 years

P(37) = 350*(1.14)^37

P(37) = 350*127.4909

P(37) = 44621.84

P(37) ≈ 44622

There would be about  44622  humpback whales in the year 2018

5 0
3 years ago
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