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Scrat [10]
2 years ago
15

An expression is given (3^2)^3•3^6/3^4

Mathematics
1 answer:
Tomtit [17]2 years ago
7 0
Hopes this if it doesn’t I any so sorry if I got it wrong:

Answer: 6561
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You roll a red, yellow and blue number cube. How many different rolls are possible?
ANEK [815]

Answer:

18 rolls are possible.

Please mark me brainliest!

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two classes are ordering pizza for a pizza party mrs.jimenez's class has 10 students and is planning to share 4 large pizzas mr.
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Each class receives the same amount of pizza because the unit rates are the same
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3 years ago
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A concert sold general admission tickets for $47.50 and lower-level seating for $97.50. The 995 tickets sold took in $68,762.50.
BabaBlast [244]

Answer: 565 general admission and 430 lower-level seating tickets sold (Answer (D))

Steps:

Let x be the number of general admission tickets, and y be the number of lower level tickets.

Set up the two equations:

68762.5 = x * 47.50 + y * 97.50       (total price as a sum of individual volumes)

x + y = 995    (sum of individual count = total count of tickets sold)

solve for x, y:

x = 995 - y

plug into the first equation:

68762.5 = (995-y) * 47.50 + y * 97.50

solve for y:

y = 430

then x = 565

x = 565 and y = 430


8 0
3 years ago
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In a sample of seven cars, each car was tested for nitrogen-oxide emissions (in grams per mile) and the following results were o
ad-work [718]

The question is incomplete. Here is the complete qeustion.

In a sample of seven cars, each car was tested for nitrogen-oxide emissions (in grams per mile) and the following results were obtained: 0.10 0.13 0.16 0.15 0.14 0.008 0.15

(a) Construct a 99% confidence interval for the mean nitrogen-oxide emissions of all cars.

(b) If the EPA requires that nitrogen-oxide emissions be less than 0.165 g/mi, based on the 99% confidence interval in (a), can we safely conclude that this requirement is being met?

Answer: (a) 0.089 ≤ μ ≤ 0.171

(b) No

Step-by-step explanation:

(a) To determine the confidence interval, first calculate the mean (X) and standard deviation (s) of the sample

X = \frac{0.1+0.13+0.16+0.15+0.14+0.08+0.15}{7}

X = 0.13

s = \sqrt{\frac{(0.1-0.13)^{2} + (0.13 - 0.13)^{2} + ... + (0.15 - 0.13)^{2}}{7-1} }

s = 0.029

The degrees of freedom is

N - 1 = 7 - 1 = 6

And since the confidence is of 99%:

α = 1 - 0.99 = 0.01

α/2 = 0.01/2 = 0.005

The t-test statistics for t_{6,0.005} is 3.707

(Value found in the t-distribution table)

Now, calculate Error:

E = t_{6,0.005} . \frac{s}{\sqrt{N} }

E = 3.707. \frac{0.029}{\sqrt{7} }

E = 0.041

The interval will be:  

0.13 - 0.041 ≤ μ ≤ 0.13+0.041

0.089 ≤ μ ≤ 0.171

(b) No, because according to the interval, the nitrode-oxide emissions range from 0.089 to 0.171, which is greater than required by EPA.

7 0
3 years ago
You sell pies at a farmers' market for $7.50 each. A group of 5 kids wants to pitch in equal1y to share one of your pies. How mu
Korolek [52]
Its 5 divided my 7.50 since each decided to pitch in to help by the pie
3 0
3 years ago
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