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Usimov [2.4K]
3 years ago
12

Find a:27=6:11, find a​

Mathematics
1 answer:
trasher [3.6K]3 years ago
3 0

Answer:

a = 14.73

Step-by-step explanation:

This ratio can also be written like this:

\frac{a}{27} = \frac{6}{11}

Then you can cross multiply to get

11a = 6 x 27

11a = 162

Divide both sides by 11

11a / 11 = 162 / 11

a = 14.73

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Simultaneous Equations <br> 2x+4y=22 <br> 2x+2y=15
Aleks04 [339]

Answer:

x = 4

y = \frac{7}{2}

Step-by-step explanation:

Ummmm

I'm not sure if this is corrrect but this is what I got

So first you solve 2x+4y=22 to find x

and you get x = -2y+11

Then you have to subtitute -2y+11 for x in 2x+2y=15

which you then get y= \frac{7}{2}

After you have to subttitute y= \frac{7}{2}  for y in x = -2y+11

which you get as x=4

Finally you get the answer as:

x = 4

y = \frac{7}{2}

5 0
3 years ago
Is 0.25 rational or irrational
love history [14]
0.25 is rational because it is a terminating (means 'its stops') fraction.
3 0
4 years ago
Identify the polygon that is the shape of the United States flag.
stellarik [79]

Answer:

d. quadrilateral

Step-by-step explanation:

3 0
3 years ago
Read 2 more answers
Someone help I don't get it
Gemiola [76]
The answer is the first choice. 
6 0
3 years ago
The owners of an amusement park selected a random sample of 200 days and recorded the number of park patrons with annual passes
Nitella [24]

Answer:

b. The method used to calculate the confidence interval has a 90% chance of producing an interval that captures the population mean number of annual pass holders in the park on any given day.

Step-by-step explanation:

The confidence interval calculated from the sample at a particular confidence level, gives a certain percentage of confidence based on the confidence level that the true mean of the population exists within the confidence interval Calculated.

For the scenario above, we can say that there is a 90% chance that the population mean number of annual pass holders in the park on a given day is within the interval (35, 51)

3 0
3 years ago
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