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Hunter-Best [27]
3 years ago
6

What is one way you can easily tell that 3/6 is not a simplified fraction

Mathematics
2 answers:
Oliga [24]3 years ago
7 0

Answer:

3/6 can be simplified

Step-by-step explanation:

3 divided by 3

6 divided by 3

it comes out to be 1/2

spin [16.1K]3 years ago
4 0
You know that 6 is divisible by 3/ 3 is a factor of 6 (3•2=6) so you could further simply 3/6 using a common factor, more specifically their Greatest Common Factor also known as the GCF
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Need help ASAP to this math question about how many gallons it will take to fill a pool
NARA [144]
1 cubic foot = 7.48 gallons

hence
\frac{7.48 gal}{1 ft^3} =1

unit conversion
\frac{3000 ft^3}{} \times \frac{7.48 gal}{1 ft^3} =  3000 \times 7.48 gal

3000 x 7.48 =22440
3 0
3 years ago
Determine the intercepts of the line.
Harlamova29_29 [7]
The y-int is (0, 0.4) and the x-int is (0.3,0).
7 0
3 years ago
g "They hired an analyst who collected a random sample of 40,361 Game of Thrones fans, and found that 8,337 of those fans said t
Viktor [21]

Answer:

The lower bound for a 90% confidence interval is 0.2033.

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of \pi, and a confidence level of 1-\alpha, we have the following confidence interval of proportions.

\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}

In which

z is the zscore that has a pvalue of 1 - \frac{\alpha}{2}.

For this problem, we have that:

n = 40361, \pi = \frac{8337}{40361} = 0.2066

90% confidence level

So \alpha = 0.1, z is the value of Z that has a pvalue of 1 - \frac{0.1}{2} = 0.95, so Z = 1.645.

The lower limit of this interval is:

\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.2066 - 1.645\sqrt{\frac{0.2066*0.7934}{40361}} = 0.2033

The lower bound for a 90% confidence interval is 0.2033.

6 0
3 years ago
What is 63 - x equals 2x + 3​
Lyrx [107]

Answer: x=20

Step-by-step explanation:

63-x=2x+3

-x-2x+63=3

-x-2x=3-63

-3x=3-63

-3x=-60

-3x/3=-60/3

X=20

please mark me brainliest!!

4 0
3 years ago
(f) how many ways are there to place a total of m indistinguishable balls into n distinguishable urns, with some urns possibly e
Blizzard [7]
The number of different ways to distribute m indistinguishable balls into n distinguishable urns is given by C(m+n-1,n-1).

For example, the number of ways 6 ball can be placed in 7 boxes is given by C(6+7-1, 7-1) = C(12, 6) = 924 ways.
4 0
3 years ago
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