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3 years ago

Equivalent ratio

Mathematics

1 answer:

Svetradugi [14.3K]3 years ago

6 0

Answer:

D.) 1:6

E.) 8:48

Step-by-step explanation:

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Allison is working two summer jobs, making $12 per hour lifeguarding and $8 per hour washing cars. Last week Allison worked 3 mo

timama [110]

Answer:

Allison worked 6 hours lifeguarding and 3 hours washing cars.

Step-by-step explanation:

Let

Number of hours Allison worked lifeguarding last week = x

Number of hours Allison worked washing cars last week = y

1. Last week Allison worked 3 more hours lifeguarding than hours washing cars hours, then

$x-y=3$

<u>Lifeguarding:</u>

$12 per hour

$12x in x hours.

<u>Washing cars:</u>

$8 pere hour

$8y in y hours.

2. Allison earned a total of $96, hence

$12x+8y=96$

You get the system of two equations:

$\left\{\begin{array}{l}x-y=3\\ \\12x+8y=96\end{array}\right.$

Plot the graphs of these two equations (see attached diagram). These line intersect at point (6,3), so Allison worked 6 hours lifeguarding and 3 hours washing cars.

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3 years ago

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

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

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Its can be divisible by 2

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