If you multiply 6 by 8.97, you get 53.82. So you would round up, therefore the answer is 54$.


- <u>We </u><u>have </u><u>given </u><u>that </u><u>the </u><u>coordinates </u><u>of </u><u>the </u><u>end </u><u>point </u><u>G </u><u>and </u><u>H </u><u>are </u><u>(</u><u> </u><u>-</u><u>6</u><u>,</u><u>5</u><u>)</u><u> </u><u>and </u><u>(</u><u> </u><u>2</u><u>,</u><u> </u><u>-</u><u>7</u><u> </u><u>)</u>

- <u>We </u><u>have </u><u>to </u><u>find </u><u>the </u><u>length </u><u>of </u><u>GH </u>

The coordinates of G = ( -6 , 5 )
The coordinates of H = ( 2 , - 7 )
<u>According </u><u>to </u><u>the </u><u>distance </u><u>formula</u><u>, </u><u> </u><u>we </u><u>get </u><u>:</u><u>-</u><u> </u>

- <u>Here</u><u>, </u><u> </u><u>x1</u><u> </u><u>=</u><u> </u><u>-</u><u>6</u><u> </u><u>,</u><u> </u><u>x2</u><u> </u><u>=</u><u> </u><u>2</u><u> </u><u>and </u><u>y1</u><u> </u><u>=</u><u> </u><u>5</u><u> </u><u>,</u><u> </u><u>y2</u><u> </u><u>=</u><u> </u><u>-</u><u>7</u>
<u>Subsitute </u><u>the </u><u>required </u><u>values </u><u>in </u><u>the </u><u>above </u><u>formula </u>








Answer:
2(3x+2) is your answer because you factor out 2. The number 2 is a factor of both 6 and 4 (2*3=6 and 2*2=4).
Answer:
p(on schedule) ≈ 0.7755
Step-by-step explanation:
A suitable probability calculator can show you this answer.
_____
The z-values corresponding to the build time limits are ...
z = (37.5 -45)/6.75 ≈ -1.1111
z = (54 -45)/6.75 ≈ 1.3333
You can look these up in a suitable CDF table and find the difference between the values you find. That will be about ...
0.90879 -0.13326 = 0.77553
The probability assembly will stay on schedule is about 78%.