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Mashutka [201]
3 years ago
11

Find y when x equals 15 if y equals 6 when x equals 30

Mathematics
1 answer:
soldier1979 [14.2K]3 years ago
6 0

X=15 and Y=?

X=30 and Y=6

By rule of three

15x6/30 = 3

The answer is Y=3
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16x+4(x-5) how do I solve this.?
Nadusha1986 [10]

Answer:20x-20

16x+4(x-5)

4(x-5)=4x-20

16x+4x

20x-20

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3 years ago
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Given f(x) = 2(4 – x)2, what is the value of F(15)? Record your answer and fat in the bubbles on your anser document​
kari74 [83]

Answer:

f(15) = 242

Step-by-step explanation:

Step 1: Define

f(x) = 2(4 - x)²

f(15) = x = 15

Step 2: Substitute and Evaluate

f(15) = 2(4 - 15)²

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3 years ago
0.4 in fraction form. Please
Digiron [165]

Answer:

4/10

Step-by-step explanation:

0.4 is said by "four-tenths." Fractions are also said the same way. Notice how there is a four and a ten. This should be easier to help you make decimals in fraction form. The four would be the numerator and the ten would be the denominator, making the final answer 4/10.

5 0
3 years ago
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While researching the cost of school lunches per week across the state, you use a sample size of 45 weekly lunch prices. The sta
Drupady [299]

We assume the lunch prices we observe are drawn from a normal distribution with true mean \mu and standard deviation 0.68 in dollars.


We average n=45 samples to get \bar{x}.


The standard deviation of the average (an experiment where we collect 45 samples and average them) is the square root of n times smaller than than the standard deviation of the individual samples. We'll write


\sigma = 0.68 / \sqrt{45} = 0.101


Our goal is to come up with a confidence interval (a,b) that we can be 90% sure contains \mu.


Our interval takes the form of ( \bar{x} - z \sigma, \bar{x} + z \sigma ) as \bar{x} is our best guess at the middle of the interval. We have to find the z that gives us 90% of the area of the bell in the "middle".


Since we're given the standard deviation of the true distribution we don't need a t distribution or anything like that. n=45 is big enough (more than 30 or so) that we can substitute the normal distribution for the t distribution anyway.


Usually the questioner is nice enough to ask for a 95% confidence interval, which by the 68-95-99.7 rule is plus or minus two sigma. Here it's a bit less; we have to look it up.


With the right table or computer we find z that corresponds to a probability p=.90 the integral of the unit normal from -z to z. Unfortunately these tables come in various flavors and we have to convert the probability to suit. Sometimes that's a one sided probability from zero to z. That would be an area aka probability of 0.45 from 0 to z (the "body") or a probability of 0.05 from z to infinity (the "tail"). Often the table is the integral of the bell from -infinity to positive z, so we'd have to find p=0.95 in that table. We know that the answer would be z=2 if our original p had been 95% so we expect a number a bit less than 2, a smaller number of standard deviations to include a bit less of the probability.


We find z=1.65 in the typical table has p=.95 from -infinity to z. So our 90% confidence interval is


( \bar{x} - 1.65 (.101),  \bar{x} + 1.65 (.101) )


in other words a margin of error of


\pm 1.65(.101) = \pm 0.167 dollars


That's around plus or minus 17 cents.




3 0
3 years ago
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Layla arked a group of students to chooue their favorite type of exerche from the choices of jogging, biling, and wwimming. The
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Answer:

50 students

Step-by-step explanation:

20 + 30 = 50

7 0
2 years ago
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