The Pythagorean Theorem ONLY works on which triangle? *

Trucks in a delivery fleet travel a mean of 110 miles per day with a standard deviation of 38 miles per day. The mileage per day

Montano1993 [528]

Answer: 0.1824

Step-by-step explanation:

Given : The mileage per day is distributed normally with

Mean : $\mu=110\text{ miles per day}$

Standard deviation : $\sigma=38\text{ miles per day}$

Let X be the random variable that represents the distance traveled by truck in one day .

Now, calculate the z-score :-

$z=\dfrac{x-\mu}{\sigma}$

For x= 132 miles per day.

$z=\dfrac{132-110}{38}\approx0.58$

For x= 159 miles per day.

$z=\dfrac{159-110}{38}\approx1.29$

Now by using standard normal distribution table, the probability that a truck drives between 132 and 159 miles in a day will be :-

$P(132$

Hence, the probability that a truck drives between 132 and 159 miles in a day =0.1824

6 0

3 years ago

I need the steps for 38

jasenka [17]

Ok so first, you put the sides into a ratio. The question says that triangle GKNM is congruent to triangle VRPT. From this, we know that side GK is congruent to VR. So, we write that as a ratio. Also, we know that KN is congruent to RP. We also write that as a ratio. Then, using equations, we solve for x. But, we aren't done yet; the expression for side KN is 3x-2 and, we already know that x is equal to 4.4
So, we do 3*4.4-2 which is 11.2. Therefore, side KN is equal to 11./2Then, the expression for side RP is x+4. We already know that x is equal to 4.4. So, we do 4.4+4 which is 8.4. Therefore, side RP is equal to 8.4.
Oh, and the scale factor is 4:3 or 1.33 because you do 8.4/6.3 ( you can also do 11.2/8.4)
hope this helps ;)

8 0

3 years ago

I really really really need help!!!!

Yakvenalex [24]

For $f$ to be continuous at $x=1$ , you need to have the limit from either side as $x\to1$ to be the same.

$\displaystyle\lim_{x\to1^-}f(x)=\lim_{x\to1^-}(|x-1|+2)=2$
$\displaystyle\lim_{x\to1^+}f(x)=\lim_{x\to1^+}(ax^2+bx)=a+b$

If $a=2$ and $b=3$ , then the limit from the right would be $2+3=5\neq2$ , so the answer to part (1) is no, the function would not be continuous under those conditions.

This basically answers part (2). For the function to be continuous, you need to satisfy the relation $a+b=2$ .

Part (c) is done similarly to part (1). This time, you need to limits from either side as $x\to2$ to match. You have

$\displaystyle\lim_{x\to2^-}f(x)=\lim_{x\to2^-}(ax^2+bx)=4a+2b$
$\displaystyle\lim_{x\to2^+}f(x)=\lim_{x\to2^+}(5x-10)=0$

So, $a$ and $b$ have to satisfy the relation $4a+2b=0$ , or $2a+b=0$ .

Part (4) is done by solving the system of equations above for $a$ and $b$ . I'll leave that to you, as well as part (5) since that's just drawing your findings.

8 0

2 years ago

The cost of catering a dinner is $11.95 per person plus $25 for delivery and setup. Which statement is true about the graph of t

valentinak56 [21]

The answer is B.) The horizontal asymptote y = 11.95 represents that the cost per person approaches $11.95 as the number of people increases.

6 0

2 years ago

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Given f(x)=3x+5 describe how the graph of g compares with the graph of f. g(x)=3(0.1)+5

mrs_skeptik [129]

The given function f(x) is: