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mrs_skeptik [129]
4 years ago
8

From 2003 to 2004 approximately how much increase in income occurred ?

Mathematics
2 answers:
igor_vitrenko [27]4 years ago
8 0

Answer:

7,000,000

Step-by-step explanation:

Elodia [21]4 years ago
4 0

Answer:

6,000,000

Step-by-step explanation:

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4/6 - 3/6 = ?
EleoNora [17]

Answer:

4/6-3/6= 1/6

Step-by-step explanation:

Hope this helps. If you need anything else I got you.

3 0
3 years ago
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Please plot the inequality on a graph y<19
VLD [36.1K]

GG easy


y<19, that means that we include all numbers greater than 19 but not including 19


so go to the graph and go up along the y axis 19 units and draw a horizontal dotted line, then shade below that

looks summat like thar attachment

7 0
4 years ago
Read 2 more answers
What could be the possible one's digit of the square root of 3721 ?
Svetllana [295]

Answer:

6,1 because 61's square root is 3721

5 0
3 years ago
primo car rental agency charges $33 per day plus $0.20 per mile. ultimo car rental agency charges $17 per day plus $0.85 per mil
belka [17]

Let's assume m represent the mileage of the car.

Given that, Primo car rental agency charges $33 per day plus $0.20 per mile.

So, the total charge for this agency = 33 + 0.20m.

Similarly total charge of the ultimo car = 17 + 0.85m.

Now we need to find the daily mileage for which the ultimo charge is four times the primo charge . So, we can set up an equation as following:

17 + 0.85m = 4 *( 33 + 0.20m )

17 + 0.85m = 132 + 0.80m By distribution property.

17 + 0.85m -0.80m = 132 + 0.80m -0.80m Subtract 0.80m from each sides.

17 + 0.05m = 132

17 + 0.05m - 17 = 132 - 17 Subtract 17 from each sides.

0.05m = 115

\frac{0.05m}{0.05} =\frac{115}{0.05} Divide each by 0.05.

So, m = 2300

Hence, the daily mileage is 2300 miles.

Hope this helps you!

3 0
3 years ago
n airline knows from experience that the distribution of the number of suitcases that get lost each week on a certain route is a
Sedbober [7]

Answer: 0.8313

Step-by-step explanation:

As per given we have,

\mu=15.5  \sigma= 3.6

Also, the distribution of the number of suitcases that get lost each week on a certain route is approximately normal.

Since , z=\dfrac{x-\mu}{\sigma}

z-score corresponds to x= 10:z=\dfrac{10-15.5}{3.6}\approx-1.53

z-score corresponds to x= 20:z=\dfrac{20-15.5}{3.6}=1.25

P-value = P(10

=P(z

Hence, the probability that during a given week the airline will lose between 10 and 20 suitcases = 0.8313

8 0
3 years ago
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