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rodikova [14]
3 years ago
14

Expand: In (4a)^3 - 3 In 4 + In a - 3 In 4 - 3 In a - 3 In 4 + 3 In a

Mathematics
1 answer:
Tpy6a [65]3 years ago
4 0

Answer:

3 ln 4 + 3 lna

Step-by-step explanation:

In (4a)^3

We know that ln x^y  = y ln x

3 ln (4a)

We also know that ln (xy) = ln x + ln y

3( ln 4 + ln a)

3 ln 4 + 3 lna

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A circle has an area of 256 square inches. If the radius of
RSB [31]

Answer:

Option C (4096π in²)

Step-by-step explanation:

Area of the circle = 256π in²

We know that Area=\pi r^{2} , where r = radius of the circle

So,

256\pi  = \pi r^{2}

Cancelling π from both the sides,

=> r^{2} = 256

=> r = \sqrt{256} = 16

The radius of the circle = 16 in.

When the radius is multiplied with 4 , new radius = 16×4 = 64 in.

New Area = \pi (64)^{2} =4096\pi \: in^{2}

6 0
3 years ago
(4x3 + 3x2 - 30x - 10) ÷ (x -3)
pogonyaev
(12x+6x-30x-10)/(x-3)=
=(-12x-10)/(x-3)
=-12x*+36- 10x+30=
=-22x+56
5 0
3 years ago
5p coins weigh approximately 3 grams.
Elza [17]

Answer: 0.3 kg

Step-by-step explanation:

£5 = 500p

500p ÷ 5 = 100

100 x 3 = 300 grams

300 ÷ 1000 = 0.3 kg

Hope this helps :)

3 0
4 years ago
Read 2 more answers
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
Can someone help me please
ki77a [65]
<span>x = -2.375 Hopefully that helps :)</span>
8 0
3 years ago
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