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MrRissso [65]
3 years ago
14

Every day, about 140,000,000 plastic water bottles are bought in the United States. Each bottle is about 0.250 meters long, and

the equator of the Earth is about 40,000,000 meters around.How many times can the Earth's equator be circled by a line of plastic bottles used in the United States every day
Mathematics
1 answer:
tatuchka [14]3 years ago
3 0
0.250*140,000,000=35,000,000
40,000,000/35,000,000= 1.14
So about 1 time 
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Determine whether the random variable is discrete or continuous. In each​ case, state the possible values of the random variable
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Answer:

A=discrete; B=continuous

Step-by-step explanation:

(a) The number of free dash throw attempts before the first shot is = discrete

​(b) The time it takes for a light bulb to burn out.=continuous

Discrete variables are countable in a finite amount of time such as the case of free dash throw attempts

Continuous Variables are more difficult to count. In statistics, time is a Continuous Variable

3 0
3 years ago
Mrs. Jackson earned a 500 bonus for signing a one-year contract to work as a nurse. Her salary is 22 per hour. If her first week
jekas [21]
Ok so your first step is to make this an equation
she gets paid x amount per hour and she works for 22 hours
so 22x plus her bonus
22x + 500 =1204
so now solve the equation
subtract both sides by 500
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-500 = -500
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22x = 704
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7 0
3 years ago
Find x+y. please help !!
Talja [164]

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Step-by-step explanation:

5 0
3 years ago
Which is the best estimate for 6,193 ÷ 48 using compatible numbers.
Gala2k [10]
It would be B. 120 you would usually round up but the closest one to 129.02083 is 130 and since there is no 130 it would be 120
8 0
3 years ago
Read 2 more answers
Prove that $5^{3^n} + 1$ is divisible by $3^{n + 1}$ for all nonnegative integers $n.$
Viktor [21]

When n=0, we have

5^{3^0} + 1 = 5^1 + 1 = 6

3^{0 + 1} = 3^1 = 3

and of course 3 | 6. ("3 divides 6", in case the notation is unfamiliar.)

Suppose this is true for n=k, that

3^{k + 1} \mid 5^{3^k} + 1

Now for n=k+1, we have

5^{3^{k+1}} + 1 = 5^{3^k \times 3} + 1 \\\\ ~~~~~~~~~~~~~ = \left(5^{3^k}\right)^3 + 1^3 \\\\ ~~~~~~~~~~~~~ = \left(5^{3^k} + 1\right) \left(\left(5^{3^k}\right)^2 - 5^{3^k} + 1\right)

so we know the left side is at least divisible by 3^{k+1} by our assumption.

It remains to show that

3 \mid \left(5^{3^k}\right)^2 - 5^{3^k} + 1

which is easily done with Fermat's little theorem. It says

a^p \equiv a \pmod p

where p is prime and a is any integer. Then for any positive integer x,

5^3 \equiv 5 \pmod 3 \implies (5^3)^x \equiv 5^x \pmod 3

Furthermore,

5^{3^k} \equiv 5^{3\times3^{k-1}} \equiv \left(5^{3^{k-1}}\right)^3 \equiv 5^{3^{k-1}} \pmod 3

which goes all the way down to

5^{3^k} \equiv 5 \pmod 3

So, we find that

\left(5^{3^k}\right)^2 - 5^{3^k} + 1 \equiv 5^2 - 5 + 1 \equiv 21 \equiv 0 \pmod3

QED

5 0
1 year ago
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