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eimsori [14]
3 years ago
6

What is the perimeter in centimeters of triangle ABC? 12 cm 14 cm 21.5 cm 35

Mathematics
1 answer:
allochka39001 [22]3 years ago
6 0
The Perimeter Answer is 12 cm
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3 years ago
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Help! Find m^KNL.<br> A. 264<br> B. 196<br> C. 184<br> D. 247
Misha Larkins [42]

Answer:

  A.  264

Step-by-step explanation:

First, we have to find the value of x. Then we can use that to find the required arc measure.

∠M = (1/2)(arc KN - arc LN)

60 = (1/2)((18x -6) -(5x +17)) = (1/2)(13x -23) . . . . substitute and simplify

120 = 13x -23 . . . . . . . multiply by 2

143 = 13x . . . . . . . . . . add 23

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5 0
4 years ago
What are the answer for your these questions?
ozzi

Answer:

<u>a. 11</u>

<u>b. 9</u>

Step-by-step explanation:

<em>A:</em>

2 x 4 = 8

8 + 3 = 11

<em>B:</em>

-2 x -2 = 4

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3 years ago
A study of 25 graduates of 4-year public colleges revealed the mean amount owed by a student in student loans was $55,051. The s
Harman [31]

Answer:

Step 1

The data represent amount.

A 90% confidence interval for the population mean is,

First, compute t-critical value then find confidence interval.

The t critical value for the 90% confidence interval is,

The sample size is small and two-tailed test. Look in the column headed and the row headed in the t distribution table by using degree of freedom is,

The t critical value for the 90% confidence interval is 1.711.

A 90% confidence interval for the population mean is .

Step 2

It is reasonable to conclude that mean of the population is actually $55000 due to a 90% confidence intrerval for population mean is between $52461.23 and $57640.77 does include $55000.

The data represent amount.

A 90% confidence interval for the population mean is,

First, compute t-critical value then find confidence interval.

The t critical value for the 90% confidence interval is,

The sample size is small and two-tailed test. Look in the column headed and the row headed in the t distribution table by using degree of freedom is,

The t critical value for the 90% confidence interval is 1.711.

A 90% confidence interval for the population mean is .

It is reasonable to conclude that mean of the population is actually $55000 due to a 90% confidence intrerval for population mean is between $52461.23 and $57640.77 does include $55000.

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5 2/3 divided by 4 this is kahnacdemy
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Answer:

answer is 17/12 (1.41)

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