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lawyer [7]
2 years ago
12

Every year the United States Department of Transportation publishes reports on the number of alcohol related and non-alcohol rel

ated highway vehicle fatalities. Below is a summary of the number of alcohol related highway vehicle fatalities from 2001 to 2010.
Line graph about Alcohol related fatalities

Determine the average number of alcohol-related fatalities from 2001 to 2006. Round to the nearest whole number.

Mathematics
1 answer:
Damm [24]2 years ago
6 0

Complete question:

The line graph relating to the question was not attached. However, the line graph has can be found in the attachment below.

Answer:

17,209

Step-by-step explanation:

The line graph provides information about alcohol-related highway fatalities between year 2001 to 2010.

Determine the average number of alcohol-related fatalities from 2001 to 2006. Round to the nearest whole number.

The average number of alcohol related fatalities between 2001 - 2006 can be calculated thus :

From the graph:

Year - - - - - - - - - - Number of fatalities

2001 - - - - - - - - - - 17401

2002 - - - - - - - - -  17525

2003 - - - - - - - - -  17013

2004 - - - - - - - - - 16694

2005 - - - - - - - - - 16885

2006 - - - - - - - - - 17738

To get the average :

Sum of fatalities / number of years

(17401 + 17525 + 17013 + 16694 + 16885 + 17738) / 6

= 103256 / 6

= 17209.333

Average number of alcohol related fatalities is 17,209 (to the nearest whole number)

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Answer: Also 57

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4 0
2 years ago
Evaluate these quantities.<br> a.−17 mod 2<br> b.144 mod 7<br> c.−101 mod 13<br> d.199 mod 19
Eduardwww [97]
Modular arithmetic is used to find the remainder when dividing the first number by the second number you can divide the term then multiply the  (or use the mod function of a graphing calculator with the first term first in the ordered pair and the second in the second term.

-17/2 = 8.5   2*.5 =1
a. 1
-144/7 = -20 5/7 . 4/7 *7 = 4
b. 4
-101/13 =
c. 3
199/19=
d.9

6 0
2 years ago
Which is a solution to (x – 3)(x + 9) = –27?
EleoNora [17]

。☆✼★ ━━━━━━━━━━━━━━  ☾  

(x - 3)(x + 9) = -27

Expand the brackets:

x^2 + 9x - 3x - 27 = -27

Simplify:

x^2 + 6x - 27 = -27

+ 27 to both sides

x^2 + 6x = 0

Factorise:

x(x + 6)

The solutions are x = 0, x = -6

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。☆✼★ ━━━━━━━━━━━━━━  ☾

5 0
2 years ago
Read 2 more answers
For the level 3 course, exam hours cost twice as much as workshop hours, workshop hours cost twice as much as lecture hours. How
natulia [17]
<h2>Answer</h2>

Cost of lectures = $7.33 per hour

<h2>Explanation</h2>

Let e the cost of the exam hours

Let w be the cost of the workshop hours

Let l be the cost of the lecture hours.

We know from our problem that exam hours cost twice as much as workshop, so:

e=2w equation (1)

We also know that workshop hours cost twice as much as lecture hours, so:

w=2l equation (2)

Finally, we also know that 3hr exams 24hr workshops  and 12hr lectures cost $528, so:

3e+24w+12l=528 equation (1)

Now, lets find the value of l:

Step 1.  Solve for l in equation (3)

3e+24w+12l=528

12l=528-3e-24w equation (4)

Step 2. Replace equation (1) in equation (4) and simplify

12l=528-3e-24w

12l=528-3(2w)-24w

12l=528-6w-24w

12l=528-30w equation (5)

Step 3. Replace equation (2) in equation (5) and solve for l

12l=528-30w

12l=528-30(2l)

12l=528-60l

72l=528

l=\frac{528}{72}

l=\frac{22}{3}

l=7.33

Cost of lectures  = $7.33 per hour



3 0
2 years ago
Read 2 more answers
2.314 divided by 1.3
kirza4 [7]
1.78 is the correct answer.
5 0
2 years ago
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