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blagie [28]
3 years ago
9

Help please! Find the length of AB:

Mathematics
2 answers:
elena-14-01-66 [18.8K]3 years ago
8 0

Answer:

Probably 16 (whatever unit)

Step-by-step explanation:

You divide both sides by three to get x after rewriting the problem.

Bezzdna [24]3 years ago
7 0

Answer:

3x+8

Step-by-step explanation:

won't it just be that because 3x+8 is between A and B

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Order the numbers from greatest to<br> least.<br> 6.057, 6.125,6.38, 6.01, 6.03 , 6.57
Sergio039 [100]

Answer:

6.57, 6.38, 6.125, 6.057, 6.03, 6.01

7 0
3 years ago
Glven the line segment RT, with endpolnt R(6,9) and<br> midpoint A(-3,7)
Rzqust [24]
It would be (-12, 5)
7 0
3 years ago
Is negative 5 a rational number
katen-ka-za [31]
Yes and rational numbers are numbers that can be written as a ratio of two integers
8 0
3 years ago
Can someone explain the process getting from:
spin [16.1K]
\rm 5=e^{3b}

The unknown b is stuck in the exponent position.
We can can fix that by using logarithms.
Log is the inverse operation of the exponential.

We'll take log of each side.
Log of what base tho?

Well, the base of our exponential is e,
so we'll take log base e of each side.

\rm log_e(5)=log_e(e^{3b})

We'll apply one of our log rules next:
\rm \log(x^y)=y\cdot\log(x)

This allows us to take the exponent out of the log,

\rm log_e(5)=(3b)log_e(e)

Another thing to remember about logs:
When the base of the log matches the inside of the log,
then the whole thing is simply 1,
\rm log_{10}(10)=1
\rm log_5(5)=1
\rm log_e(e)=1

So our equation simplifies to this,

\rm log_e(5)=(3b)\cdot1

As a final step, divide both sides by 3,

\rm \frac13log_e(5)=b

k, hope that helps!


6 0
3 years ago
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
Read 2 more answers
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