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2 years ago

12

For the triangle, select the correct order of the side lengths from least to greatest.

1 answer:

Andre45 [30]2 years ago

5 0

M<j then m<l then m<k

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The cost c, for manufacturing x flags is given by c=x? -24x +39. Find the number of flags manufactured if the cost is $120.

IRINA_888 [86]

9514 1404 393

Answer:

27

Step-by-step explanation:

You want to find x when c = 120.

120 = x^2 -24x +39

x^2 -24x -81 = 0 . . . . . . subtract 120 to put in standard form

(x -27)(x +3) = 0 . . . . . .factor it

Solutions are ...

x = 27, x = -3 . . . . . values of x that make the factors zero

The number of flags manufactured for a cost of $120 is 27 flags.

8 0

2 years ago

Help pleaseeeeeeeeeee

avanturin [10]

I believe your answer is going to be increasing linear. If it's doubling constantly then it's increasing so that takes out 2 options. And since it's constant, it isn't exponential, therefore you only have increasing linear left.

I hope I was able to help. Best of luck.

5 0

2 years ago

a cube is dilated by a scale factor of 3/4 to create a new cube. what is the surface area of the new cube.

jeka57 [31]

Answer:

9/16

Step-by-step explanation:

i took the test ;)

7 0

2 years ago

Read 2 more answers

The probability density function of the time you arrive at a terminal (in minutes after 8:00 A.M.) is f(x) = 0.1 exp(−0.1x) for

Blababa [14]

$f_X(x)=\begin{cases}0.1e^{-0.1x}&\text{for }x>0\\0&\text{otherwise}\end{cases}$

a. 9:00 AM is the 60 minute mark:

$f_X(60)=0.1e^{-0.1\cdot60}\approx0.000248$

b. 8:15 and 8:30 AM are the 15 and 30 minute marks, respectively. The probability of arriving at some point between them is

$\displaystyle\int_{15}^{30}f_X(x)\,\mathrm dx\approx0.173$

c. The probability of arriving on any given day before 8:40 AM (the 40 minute mark) is

$\displaystyle\int_0^{40}f_X(x)\,\mathrm dx\approx0.982$

The probability of doing so for at least 2 of 5 days is

$\displaystyle\sum_{n=2}^5\binom5n(0.982)^n(1-0.982)^{5-n}\approx1$

i.e. you're virtually guaranteed to arrive within the first 40 minutes at least twice.

d. Integrate the PDF to obtain the CDF:

$F_X(x)=\displaystyle\int_{-\infty}^xf_X(t)\,\mathrm dt=\begin{cases}0&\text{for }x$

Then the desired probability is

$F_X(30)-F_X(15)\approx0.950-0.777=0.173$

7 0

3 years ago

(1) Tom took a trip of 1,020 miles. He traveled by train at 55 miles an hour and the same number of hours by plane at 285 mph.

Maurinko [17]

<h3>Answer:</h3>

B) 6 hours

<h3>Step-by-step explanation:</h3>

Let h represent the total trip time. Then h/2 is the time spent traveling at each speed. The distance covered is ...

... distance = speed · time

The sum of the distances in each mode is the total distance.

... 1020 = 55·h/2 +285·h/2

... 2040 = h·(55+285) . . . . multiply by 2

... 2040/340 = h = 6 . . . . . . . divide by the coefficient of h

The trip took a total of 6 hours.

4 0

2 years ago