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Studentka2010 [4]
3 years ago
6

FIND THE UNIT RATE AND ROUND TO THE NEAREST HUNDREDTHS IF NECESSARY. Jasigreen's class opened last year and serviced 6,840 custo

mers in 45 days.
Mathematics
1 answer:
Vladimir [108]3 years ago
4 0
<span>6840 customers ... 45 days<span>
x customers = ? ... 1 day

If you would like to know how many customers were in Jasigreen's class in 1 day, you can calculate this using the following steps:

6840 * 1 = 45 * x
6840 = 45 * x     /45
x = 6840 / 45
x = 152 customers

<span>The correct result would be </span>152 customers<span>.</span></span></span>
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Which expression is equivalent to ( 4mn / m^-2n^6)^-2
inn [45]

Here are a few rules with exponents that apply here:

  1. Raising a power to a power: (x^m)^n=x^{m*n}
  2. Dividing exponents of the same base: \frac{x^m}{x^n}=x^{m-n}
  3. Converting negative exponents to positive ones: x^{-m}=\frac{1}{x^m}\ \textsf{and}\ \frac{1}{x^{-m}}=x^m

Firstly, solve the outer exponent:

(\frac{4mn}{m^{-2}n^{6}})^{-2}=\frac{4^{-2}m^{-2}n^{-2}}{m^{-2*-2}n^{6*-2}}=\frac{4^{-2}m^{-2}n^{-2}}{m^4n^{-12}}

Next, divide:

\frac{4^{-2}m^{-2}n^{-2}}{m^4n^{-12}}=4^{-2}m^{-2-4}n^{-2-(-12)}=4^{-2}m^{-6}n^{10}

Next, convert the negative exponents:

4^{-2}m^{-6}n^{10}=\frac{n^{10}}{4^2m^6}=\frac{n^{10}}{16m^6}

<u>Your final answer is n^10/16m^6 , or B.</u>

8 0
3 years ago
Determine whether the data described are qualitative or quantitative. The maximum speed limit on interstate highways.
dalvyx [7]
The answer is B: Quantitative
6 0
3 years ago
The vertex of the parabola below is at the point (1.-3). Which of the equations below could be the one for this parabola?
Ilia_Sergeevich [38]

Answer: Choice D. y = (x-1)^2 - 3

The vertex is (h,k) = (1,-3). So h = 1 and k = -3.

We have a = 1 as the leading coefficient.

Plug those values into the equation below

y = a(x-h)^2 + k

y = 1(x - 1)^2 + (-3)

y = (x - 1)^2 - 3

3 0
3 years ago
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Suppose a basketball player has made 390 out of 434 free throws. If the player makes the next 2 free throws, I will pay you $40.
alukav5142 [94]

Answer: a) -$0.19, b) -$111.72 .

Step-by-step explanation:

Since we have given that

Number of free throws = 434

Number of throws made by them = 390

Amount for making the next 2 free throws = $40

Amount otherwise he has to pay = $169

a) Find the expected value of the proposition.

Expected value of success in next 2 free throws = \dfrac{390}{434}\times \dfrac{391}{435}=0.8077

Expected value would be

0.8077\times 40+(1-0.8077)\times -169\\\\=32.308-32.4987\\\\=-\$0.19

b)  If you played this game 588 times how much would you expect to win or lose?

Number of times they played the game = 588

So, Expected value would be

588\times -0.19\\\\=-\$111.72

Hence, a) -$0.19, b) -$111.72

8 0
3 years ago
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ValentinkaMS [17]

98.43

May nilbin be with you

5 0
3 years ago
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