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Verdich [7]
3 years ago
13

Drag each equation to the correct location on the table.

Mathematics
1 answer:
11Alexandr11 [23.1K]3 years ago
5 0

Answer:

A) 4 + 3a + 8 < 16    ⇒a < 4/3

B) 5a − 4a + 6 < 8    ⇒a < 2

C) 2a + 3 < 9 − 3 a  ⇒ a < 6/5

D) 14 + 5a < 12a + 1     ⇒ a > 13 / 7

E) 2a + 3 > 10 − 3a   ⇒ a < 7/5

F) 3a − 2a + 1 > 2  ⇒ a < 1

Step-by-step explanation:

The given question is INCOMPLETE as the options are MISSING.

So, here let us solve each inequality for the value of a.

A) 4 + 3a + 8 < 16

⇒ 3 a + 12  < 16

or, 3 a < 16 -12  = 4

or, a < 4/3

B) 5a − 4a + 6 < 8

⇒ a < 8 - 6

or, a < 2

C) 2a + 3 < 9 − 3a

⇒ 2a +  3a  < 9 - 3

or, 5a < 6 , or a < 6/5

D) 14 + 5a < 12a + 1

⇒ 14  - 1 <12 a  - 5 a

or, 13 < 7 a

or, a > 13 / 7

E) 2a + 3 > 10 − 3a

⇒ 2a +  3a  < 10 - 3

or, 5a < 7 , or a < 7/5

F) 3a − 2a + 1 > 2

⇒ a  < 2 - 1

or, a < 1

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Step-by-step explanation:

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3 years ago
A healthcare provider monitors the number of CAT scans performed each month in each of its clinics. The most recent year of data
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Answer: a. 1.981 < μ < 2.18

              b. Yes.

Step-by-step explanation:

A. For this sample, we will use t-distribution because we're estimating the standard deviation, i.e., we are calculating the standard deviation, and the sample is small, n = 12.

First, we calculate mean of the sample:

\overline{x}=\frac{\Sigma x}{n}

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Now, we estimate standard deviation:

s=\sqrt{\frac{\Sigma (x-\overline{x})^{2}}{n-1} }

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For t-score, we need to determine degree of freedom and \frac{\alpha}{2}:

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The 95% two-sided CI on the mean is 1.981 < μ < 2.18.

B. We are 95% confident that the true population mean for this clinic is between 1.981 and 2.18. Since the mean number performed by all clinics has been 1.95, and this mean is less than the interval, there is evidence that this particular clinic performs more scans than the overall system average.

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