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Gwar [14]
3 years ago
9

PLEASE HELP!!!!! LOOK AT PHOTO

Mathematics
2 answers:
dybincka [34]3 years ago
5 0

Answer:

I dont understand the question -_-

Korvikt [17]3 years ago
5 0

Answer:

I not sure

But I think is 33.3

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2x - 4y = 10
Bas_tet [7]

Answer:

Substitution, but this system of equations has no solution.

Step-by-step explanation:

2x-4y=10

2y+6=x

----------------

2(2y+6)-4y=10

4y+12-4y=10

12=10

no solution.

5 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
Help me on this one too please.
Kobotan [32]
The answer is B I did something like this before I believe that is the answer
4 0
3 years ago
Read 2 more answers
Can someone write a story with dialogue about a kid who gets to be President of the United States for the day.?
elena55 [62]
How many words will it consist of
6 0
4 years ago
Read 2 more answers
Robin has 208 beads to share equally with 8 friends. Which equation can Robin use to find how many beads each friend will get? P
antoniya [11.8K]

Answer:

208/8=26

Step-by-step explanation:

Each friend would get 26 beads.

7 0
3 years ago
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