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Black_prince [1.1K]
3 years ago
12

Suppose the linear regression line y = 4.009x - 77.531 predicts a pizza

Mathematics
1 answer:
fredd [130]3 years ago
7 0

Answer:

Option C, 1225

Step-by-step explanation:

y=4.009x-77.531

selling 325 pizzas, so x=325, now putting the value of x in the equation,

y=4.009×325-77.531

or, y=1225.394

y = 1225.394, which is the profit, if we approximate it to the nearest value, it's $1225

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19 yards divided by 2.5 minutes.

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Can someone please help me with this one question
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Solve for d<br> 20(d − 4) + 4d ≤ 8
tatiyna

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Step-by-step explanation:

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Are these two lines perpendicular?
lys-0071 [83]

Answer:

Step-by-step explanation:

This is the image of the graph you determine if that's perpendicular.

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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
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