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Alik [6]
2 years ago
10

The graph above shows the distance that a truck travels as a function of the total number of liters of fuel used. The first 6 li

ters of fuel were used for city driving, while the next 10 liters were used for highway driving. Based on the graph, how does the average miles per liter for city driving compare to the average miles per liter for highway driving?​

Mathematics
1 answer:
4vir4ik [10]2 years ago
6 0

Answer:

The city and highway average are about the same

Step-by-step explanation:

The m/slope of this graph tells us the relation between fuel used and miles traveled.

So, by looking at the slope, we can tell the average miles per liter for driving.

Because the slope of this graph stays fairly consistent (the slope stays the same throughout the entire graph, and there is no sudden change in the miles per liter ratio), we know that the fuel usage does not change.

So, we don't even have to calculate the slope for each separate section, we know that the usage is about the same (from simply looking at the graph).

(Also, if you specifically wanted to calculate slope to confirm, like I just did, you could do this too. Without even having to calculate the slope itself, you can find one point in each section to judge. The points I used were (2, 10) [10m traveled with 2 liters, meaning 5m traveled with 1 liter] and (10, 50) [50m traveled with 10 liters, 5m traveled with 1 liter])

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Answer:

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A, b, c and d are positive integers, such that a+b+ ab = 76, c+d+ cd = 54. Find (a+b+c+d)·a·b·c·d.
lutik1710 [3]

Notice that

(1 + <em>x</em>)(1 + <em>y</em>) = 1 + <em>x</em> + <em>y</em> + <em>x y</em>

So we can add 1 to both sides of both equations, and we use the property above to get

<em>a</em> + <em>b</em> + <em>a b</em> = 76  ==>  (1 + <em>a</em>)(1 + <em>b</em>) = 77

and

<em>c</em> + <em>d</em> + <em>c d</em> = 54  ==>  (1 + <em>c</em>)(1 + <em>d</em>) = 55

Now, 77 = 7*11 and 55 = 5*11, so we get

<em>a</em> + 1 = 7  ==>  <em>a</em> = 6

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(or the other way around, since the given relations are symmetric)

and

<em>c</em> + 1 = 5  ==>  <em>c</em> = 4

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Two events E1 and E2 are called independent if p(E1 â© E2) = p(E1)p(E2). For each of the following pairs of events, which are su
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Answer:  a) Independent

b) Independent

c) Dependent

Step-by-step explanation:

Since, If a coin is tossed three times,

Then, total number of outcomes, n(S) = 8

a)  E_1 : tails comes up with the coin is tossed the first time;

E_1 = { TTT, THH, THT, TTH }

E_2 :  heads comes up when the coin is tossed the second time.

E_2 = { THT, HHH, THH, HHT }

Thus, n(E_1)=4

⇒  P(E_1)=\frac{n(E_1)}{n(S)}=\frac{4}{8}=\frac{1}{2}

Similarly,  P(E_2)=\frac{1}{2}

⇒  P(E_1)\times P(E_2)=\frac{1}{2}\times \frac{1}{2}=\frac{1}{4}

Since,   E_1\cap E_2 = { THH, THT }

n(E_1\cap E_2) = 2

⇒ P(E_1\cap E_2) = \frac{n(E_1\cap E_2)}{n(S)}= \frac{2}{8}=\frac{1}{4}

Thus,  P(E_1\cap E_2)=P(E_1)\timesP(E_2)

Therefore,  E_1 and E_2 are independent events.

B)  E_1 :  the first coin comes up tails

E_1 = { TTT, THH, THT, TTH }

E_2 :   two, and not three, heads come up in a row

E_2 = { HHT, THH }

Thus, n(E_1)=4

⇒  P(E_1)=\frac{n(E_1)}{n(S)}=\frac{4}{8}=\frac{1}{2}

Similarly,  P(E_2)=\frac{1}{4}

⇒  P(E_1)\times P(E_2)=\frac{1}{2}\times \frac{1}{4}=\frac{1}{8}

Since,   E_1\cap E_2 = { THH }

n(E_1\cap E_2) = 1

⇒ P(E_1\cap E_2) = \frac{n(E_1\cap E_2)}{n(S)}= \frac{1}{8}

Thus,  P(E_1\cap E_2)=P(E_1)\timesP(E_2)

Therefore,  E_1 and E_2 are independent events.

C)  E_1 :  the second coin comes up tails;

E_1 = { HTH, HTT, TTT, TTH }

E_2 :   two, and not three, heads come up in a row

E_2 = { HHT, THH }

Thus, n(E_1)=4

⇒  P(E_1)=\frac{n(E_1)}{n(S)}=\frac{4}{8}=\frac{1}{2}

Similarly,  P(E_2)=\frac{1}{4}

⇒  P(E_1)\times P(E_2)=\frac{1}{2}\times \frac{1}{4}=\frac{1}{8}

Since,   E_1\cap E_2 = \phi

n(E_1\cap E_2) = 0

⇒ P(E_1\cap E_2) = 0

Thus,  P(E_1\cap E_2)\neq P(E_1)\timesP(E_2)

Therefore,  E_1 and E_2 are dependent events.


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