All categories

3 years ago

A career placement test eliminates a profession when a person receives a score of 18 or less. If the score is calculated

Mathematics

2 answers:

Volgvan3 years ago

3 0

Answer:

See the figure attached

Step-by-step explanation:

Let's call :

x: positive response

y: negative response

then the score is:

2x - 3y

A career placement test eliminates a profession when a person receives a score of 18 or less, that's represented by the following inequality:

2x - 3y ≤ 18

lubasha [3.4K]3 years ago

3 0

Answer:

Step-by-step explanation:

You might be interested in

In 4 month, 104 gallons were used(determine gallons per month)

saw5 [17]

All you have to do is divide 104 by 4! The awnser is 26

6 0

3 years ago

Given the equation below, what is the value of 3 + 6(79 – s)? 7= -s+ 82

ss7ja [257]

Answer:

1 expression) 477 - 6s

2 equation) s = 75

Step-by-step explanation:

Also, the first expression is simplified

4 0

3 years ago

Math word problem. What's the equation I use?

UkoKoshka [18]

You would have to find the proportionality of the two. If his shadow is 3 feet long, and the tree's shadow is 9 that means it is proportioned by 3. So the height of the tree would be 15.

3 0

3 years ago

If the altitude of an equilateral triangle is 9 what is the perimeter

Volgvan

The angles in an equilateral triangle are 60 degrees.

sin 60 = 9 / h where h is hypotenuse ( = side of the triangle)

h = 9 / sin 60 = 10.392

Perimeter = 3 * h = 31.18 to nearest hundredth

3 0

3 years ago

Read 2 more answers

Changes in airport procedures require considerable planning. Arrival rates of aircrft are important factors that muct be taken i

Irina18 [472]

Answer:

a) 0.1558

b) 0.7983

c) 0.1478

Step-by-step explanation:

If we suppose that small aircraft arrive at the airport according to a Poisson process at the rate of 5.5 per hour and if X is the random variable that measures the number of arrivals in one hour, then the probability of k arrivals in one hour is given by:

$\bf P(X=k)=\displaystyle\frac{(5.5)^ke^{-5.5}}{k!}$

(a) What is the probability that exactly 4 small aircraft arrive during a 1-hour period?

$\bf P(X=4)=\displaystyle\frac{(5.5)^4e^{-5.5}}{4!}=0.1558$

(b) What is the probability that at least 4 arrive during a 1-hour period?

$\bf P(X\geq4)=1-P(X$

(c) If we define a working day as 12 hours, what is the probability that at least 75 small aircraft arrive during a working day?

If we redefine the time interval as 12 hours instead of one hour, then the rate changes from 5.5 per hour to 12*5.5 = 66 per working day, and the pdf is now

$\bf P(X=k)=\displaystyle\frac{(66)^ke^{-66}}{k!}$

and we want P(X ≥ 75) = 1-P(X<75). But

$\bf P(X$

hence

P(X ≥ 75) = 1-0.852 = 0.1478

7 0

3 years ago