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insens350 [35]
2 years ago
12

What is the roasting time for a 12-pound turkey?

Mathematics
1 answer:
NeX [460]2 years ago
5 0
The roasting time for a 12 pound turkey is 3 hours
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72 + 8x + 18x = 36x
ArbitrLikvidat [17]
72 + 8x + 18x = 36x
 

72+26x=36x
 

72=36x-26x
  \text{Subtract 26x from both sides} 

72=10x
 

\text{Divide both sides by 10} 

x=36/5
4 0
3 years ago
Calculus AB
wariber [46]
Assuming you mean
\lim_{x \to 4} \frac{\sqrt{x+5}-3}{4-x}

that means as x approaches 4

if we sub 0 for x we get
0/0
and intermitent form
use l'hopital's rule

so
take the derivitive of the top and bottom seperatly
l'hopitals rule is something like
if \lim_{x \to n} \frac{f(x)}{g(x)} results in 0/0 or -∞/∞ or∞/∞ then keep doing it until f(n)/g(n) gives a form that isn't intermitent

so

take derivitive of top and bottom
\dfrac{\frac{1}{2\sqrt{x+5}}}{-1}
now, if we subsitute 4 for x we get
\dfrac{\frac{1}{2\sqrt{4+5}}}{-1}=

\dfrac{\frac{1}{2\sqrt{9}}}{-1}=

\dfrac{\frac{1}{2(3)}}{-1}=

\dfrac{\frac{1}{6}}{-1}=

\dfrac{1}{-6}=\dfrac{-1}{6}
8 0
3 years ago
Find mAC<br><br> Help me please
aleksandrvk [35]

Answer:

130 degrees

Step-by-step explanation:

arc CA is 50 since the angle is 50 and then you would subtract from 180 since it is half the circle which leaves you with C, or 130 degrees.

5 0
2 years ago
Which of the following points is not collinear with (3, 6) and (-2, -9)?
Lilit [14]
Its B, cause it’s collinear
5 0
3 years ago
Read 2 more answers
A cellular phone company monitors monthly phone usage. The following data represent the monthly phone use in minutes of one part
Sergeu [11.5K]

Answer:

The standard deviation increased but there was no change in the interquantile range          

Step-by-step explanation:

We are given the following data in the question:

320, 411, 348, 537, 420, 449, 462, 403, 454, 517, 515, 358, 438, 541, 387, 368, 502, 437, 431, 428.

n = 20

a) Formula:

\text{Standard Deviation} = \sqrt{\displaystyle\frac{\sum (x_i -\bar{x})^2}{n-1}}  

where x_i are data points, \bar{x} is the mean and n is the number of observations.  

Mean = \displaystyle\frac{\text{Sum of all observations}}{\text{Total number of observation}}

Mean =\displaystyle\frac{8726}{20} = 436.3

Sum of squares of differences =

13525.69 + 640.09 + 7796.89 + 10140.49 + 265.69 + 161.29 + 660.49 + 1108.89 + 313.29 + 6512.49 + 6193.69 + 6130.89 + 2.89 + 10962.09 + 2430.49 + 4664.89 + 4316.49 + 0.49 + 28.09 + 68.89 = 75924.2

S.D = \sqrt{\frac{75924.2}{19}} = 63.21

Sorted Data = 320, 348, 358, 368, 387, 403, 411, 420, 428, 431, 437, 438, 449, 454, 462, 502, 515, 517, 537, 541

IQR = Q_3 - Q_1\\Q_3 = \text{upper median},\\Q_1 = \text{ lower median}

Median:\\\text{If n is odd, then}\\\\Median = \displaystyle\frac{n+1}{2}th ~term \\\\\text{If n is even, then}\\\\Median = \displaystyle\frac{\frac{n}{2}th~term + (\frac{n}{2}+1)th~term}{2}

Median = \frac{431 + 437}{2} = 434

Q_1 = \frac{387 + 403}{2} = 395\\\\Q_3 = \frac{462 + 502}{2} = 482

IQR = 482 - 395 = 87

b) After changing the observation

0, 411, 348, 537, 420, 449, 462, 403, 454, 517, 515, 358, 438, 541, 387, 368, 502, 437, 431, 428

Mean =\displaystyle\frac{8406}{20} = 420.3

Sum of squares of differences =

176652.09 + 86.49 + 5227.29 + 13618.89 + 0.09 + 823.69 + 1738.89 + 299.29 + 1135.69 + 9350.89 + 8968.09 + 3881.29 + 313.29 + 14568.49 + 1108.89 + 2735.29 + 6674.89 + 278.89 + 114.49 + 59.29 = 247636.2

S.D = \sqrt{\frac{247636.2}{19}} = 114.16

Sorted Data = 0, 348, 358, 368, 387, 403, 411, 420, 428, 431, 437, 438, 449, 454, 462, 502, 515, 517, 537, 541

Median = \frac{431 + 437}{2} = 434

Q_1 = \frac{387 + 403}{2} = 395\\\\Q_3 = \frac{462 + 502}{2} = 482

IQR = 482 - 395 = 87

Thus. the standard deviation increased but there was no change in the interquantile range.

5 0
3 years ago
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