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pantera1 [17]
2 years ago
6

Lashonda exercises at least 50 minutes per day. Use t to represent Lashonda's amount of exercise (in minutes per day).

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

The question is an illustration of inequalities.

The inequality that represents Lashonda's amount of exercise is: \mathbf{t \ge 50}

The given parameter is:

\mathbf{Minutes = At\ least\ 50}

In inequality, at least means <em>greater than or equal to</em>.

So, the above equation becomes:

\mathbf{Minutes \ge 50}

From the question, we understand that the number of minutes should be represented with t.

So, we have:

\mathbf{t \ge 50}

Hence, the inequality is: \mathbf{t \ge 50}

Read more about inequalities at:

brainly.com/question/15137133

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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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