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Bingel [31]
3 years ago
14

According to the line plot, how much more is the total weight of the 1/2 of a pound dog toys than the weight of the 5/8 of a pou

nd dog toys?
Mathematics
1 answer:
USPshnik [31]3 years ago
6 0
You didn’t attach the line plot but I’ll do my best.

1 pound = 16 ounces

It says of a pound dog toys, meaning that we will have to do multiplication:

1/2 * 1 pound

= 1/2 * 16 ounces

= 8 ounces

5/8 * 16 ounces

= 10

Next, find the difference:

10 - 8 = 2

The total weight of 5/8 of a pound of dog toys is 2 ounces greater than 1/2 of a pound of dog toys.


Also, the last bit of the question is worded incorrectly. 5/8 > 1/2 right off the bat. I’m not sure who designed the question but I think you should let them know since it would likely confuse individuals who encounter this question. Hope this helps!
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A bus travels on an east-west highway connecting two cities A and B that are 100 miles apart. There are 2 services stations alon
melamori03 [73]

Answer:

51/4

Step-by-step explanation:

To begin with you have to understand what is the distribution of the random variable. If X represents the point where the bus breaks down. That is correct.  

X~ Uniform(0,100)

Then the probability mass function is given as follows.

f(x) = P(X=x) = 1/100  \,\,\,\, \text{if} \,\,\,\, 0 \leq x \leq 100\\f(x) = P(X=x) = 0  \,\,\,\, \text{otherwise}

Now, imagine that the D represents the distance from the break down point to the nearest station. Think about this, the first service station is 20 meters away from city A, and the second station is located  70 meters away from city A then the mid point between 20 and 70  is (70+20)/2 = 45 then we can represent D as follows

D(x) =\left\{ \begin{array}{ll}  x  & \mbox{if } 0\leq x \leq 20 \\  x-20 & \mbox{if } 20\leq x < 45\\                70-x & \mbox{if } 45 \leq x \leq 70\\                x-70 & \mbox{if } 70 \leq x \leq 100\\ \end{array}\right.

Now, as we said before X represents the random variable where the bus breaks down, then we form a new random variable Y = D(X), Y is a random variable as well, remember that there is a theorem that says that

E[Y] = E[D(X)] = \int\limits_{-\infty}^{\infty} D(x) f(x) \,\, dx

Where f(x) is the probability mass function of X. Using the information of our problem

E[Y] = \int\limits_{-\infty}^{\infty}  D(x)f(x) dx \\= \frac{1}{100} \bigg[ \int\limits_{0}^{20} x dx +\int\limits_{20}^{45} (x-20) dx +\int\limits_{45}^{70} (70-x) dx +\int\limits_{70}^{100} (x-70) dx  \bigg]\\= \frac{51}{4} = 12.75

3 0
3 years ago
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