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muminat
4 years ago
14

Please help. I'm stuck.

Mathematics
1 answer:
Bezzdna [24]4 years ago
4 0
The root \sqrt{10} can be converted into the power 10^{ \frac{1}{2} }. Therefore we can rewrite the problem as (10^{ \frac{1}{2} } )^ {\frac{3}{4} x} and then follow the exponent rules about a power to a power, multiplying 1/2 and 3/4 together.

Thus the problem becomes 10^{\frac{3}{8}x}, which then can be turned into \sqrt[8]{10} ^{3x}, making the last option our answer.

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The answer would be 12:7 . You have to change feet into inches and calculate it from there . Doing that would get you 48:28 and if you put it into simplest form it'd be 12:7 dividing both by 4.
6 0
3 years ago
The ideal (daytime) noise-level for hospitals is 45 decibels with a standard deviation of 10 db; which is to say, this may not b
WINSTONCH [101]

Answer:

<em>(a) A 99% confidence interval for the actual mean noise level in hospitals is </em><em>(44.02 db, 49.98 db)</em><em>. </em>

<em>(b) We can be 90% confident that the actual mean noise level in hospitals is </em><em>47 db</em><em> with a margin of error of </em><em>1.89 db</em><em>. </em>

<em>(c) Unless our sample (of 81 hospitals) is among the most unusual 2% of samples, the actual mean noise level in hospitals is between </em><em>44.41 db and 49.59 db</em><em>. </em>

<em />

Step-by-step explanation:

<em>The problem is incomplete. The questions are:</em>

<em />

<em>(a) A 99% confidence interval for the actual mean noise level in hospitals is </em><em>(44.02 db, 49.98 db)</em><em>. </em>

For a 99% CI, the value of z is z=2.58

Then, the confidence interval for the mean is:

M-z\sigma/\sqrt{n}\leq\mu\leq M-z\sigma/\sqrt{n}\\\\47-2.58*10/\sqrt{75}  \leq\mu\leq47+2.58*10/\sqrt{75}\\\\47-2.98\leq\mu\leq47+2.98\\\\44.02\leq\mu\leq 49.98

<em>(b) We can be 90% confident that the actual mean noise level in hospitals is </em><em>47 db</em><em> with a margin of error of </em><em>1.89 db</em><em>. </em>

For a 90% CI, the value of z is z=1.64.

Then, we can calculate the margin of error as:

E=z*\sigma/\sqrt{n}=1.64*10/\sqrt{75}=1.89

<em>(c) Unless our sample (of 81 hospitals) is among the most unusual 2% of samples, the actual mean noise level in hospitals is between </em><em>44.41 db and 49.59 db</em><em>. </em>

The 2% tails data corresponds, in the standard normal distirbution, to the values of z whose absolute value is higher than 2.33.

The values of db for these critical values are:

X_1=M+z_1*\sigma/\sqrt{n}=47+(-2.33)*10/\sqrt{81}=47-2.59=44.41\\\\\\ X_2=M+z_2*\sigma/\sqrt{n}=47+(2.33)*10/\sqrt{81}=47+2.59=49.59

3 0
4 years ago
In (x+1)- In x=2 ?????
Marta_Voda [28]

Answer:

this needs to be worded different in order to understand it

Step-by-step explanation:

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3 years ago
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Up and down is vertical. At w = 0 there's a vertical asymptote.
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The SAT mathematics scores in the state of Florida for this year are approximately normally distributed with a mean of 500 and a
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99.7%

This is because the likelihood of a point being within 2 SDs of the mean is 99.7%. 
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