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dangina [55]
1 year ago
13

A company administers an "aptitude test for managers" to aid in selecting new management trainees. Prior experience suggests tha

t 60 percent of all applications for management trainee positions would be successful if they were hired. Furthermore, past experience with the aptitude test indicates that 85 percent of applicants who turn out to be successful managers pass the test and 90 percent of applicants who do not turn out to be successful managers fail the test.
Mathematics
1 answer:
RoseWind [281]1 year ago
5 0

Answer:

0.94

Step-by-step explanation:

The question after this basically is:

<em>"If the applicant passes the "aptitude test for managers", what is the probability that the applicant will succeed in the management position?"</em>

<em />

So,

P(successful if hired) = 60% = 0.6 [let it be P(x)]

P(success at passing the test) = 85% = 0.85   [let it be P(y)]

P(successful and pass the test) =  P(x) + P(y) -[P(x)*P(y)]

So,

P(successful and pass the test) = 0.6 + 0.85 - (0.6*0.85) = 0.94 (94%)

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1 year ago
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A boy had 20 cents. He bought x pencils for 3 cents each. If y equals the number of pennies left, write an equation showing the
Vesna [10]

Hey there!!

The total number of cents - 20

Cost for each pencil - 3 cents

Number of pencils bought - ' x '

y = number of pennies left

What is the domain ?

Show how y is dependent on x ..

Let's get this into an equation :

... The total cost for x pencils bought = 3x

... Number of pennies left = 20 - 3x

... y = number of pennies left

... y = 20 - 3x

Notice : If the x value changes, the y value changes too

... If x = 1 , then , y = 17; If x = 2 , then , y = 14

Hence, we could say y is dependent upon x

Domain = ?

... Remember - The total number of pennies = 20 ; hence, the total cost cannot go above 20 cents.

Hence, we will have to work with inequalities

The equation  :

... 20 ≥ 3x

Divide 3 on both sides

20 / 3 ≥ x

x ≥ 6.667

Let's take this as 6

Hence , the domain will be :

D : { 1 , 2 , 3 , 4 , 5 , 6 }

Hope my answer helps!

6 0
2 years ago
A triangle ABC is inscribed in a circle, such that AB is a diameter. What are the measures of angles of this triangle if: measur
shutvik [7]

Answer:

The measures of angles of this Δ are 23° , 67° , 90°

Step-by-step explanation:

* Lets talk about some facts in the circle

- An inscribed angle is an angle made from points sitting on the

 circle's circumference

- A central angle is the angle formed when the vertex is at the center

 of the circle

- The measure of an arc of a circle is equal to the measure of the

 central angle that intercepts the arc.

- The measure of an inscribed angle is equal to 1/2 the measure of

 its intercepted arc

- An angle inscribed across a circle's diameter is always a right angle

- The triangle is inscribed in a circle if their vertices lie on the

  circumference of the circle, and their angles will be inscribed

  angles in the circle

* Now lets solve the problem

- Δ ABC is inscribed in a circle

∵ its side AB is a diameter of the circle

∵ Its vertex C is on the circle

∴ ∠C is inscribed and across the circle's diameter

∴ ∠C is a right angle

∴ m∠C = 90°

∵ The measure of arc BC = 134°

∵ ∠A is inscribed angle subtended by arc BC

∵ The measure of an inscribed angle is equal to 1/2 the measure

   of its intercepted arc

∴ m∠A = 1/2 × 134° = 67°

∵ The sum of the measures of the interior angles of a triangle is 180°

∵ m∠A = 67°

∵ m∠C = 67°

∵ m∠A + m∠B + m∠C = 180°

∴ 67° + m∠B + 90° = 180°

∴ 157° + m∠B = 180° ⇒ subtract 157 from both sides

∴ m∠B = 23°

* The measures of angles of this Δ are 23° , 67° , 90°

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1 year ago
What is 6 divided by 612
kondaur [170]
612/6=102
check: 6x102=612

Hope this helped! :))
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Let Xn be the random variable that equals the number of tails minus the number of heads when n fair coins are flipped. What is t
Firlakuza [10]

Answer:

the expected value of Xn , E(Xn) = 0 and the variance σ²(Xn) = n*(1-2n)

Step-by-step explanation:

If X1= number of tails when n fair coins are flipped , then X1 follows a binomial distribution with E(X1) = n*p , p=0,5 and the number of heads obtained is X2=n-X1

therefore

Xn =X1-X2 = X1- (n-X1) = 2X1-n

thus

E(Xn) =∑ (2*X1-n) p(X1) =  2*∑[X1 p(X1)] -n∑p(X1) = 2*E(X1)-n = 2*n*p--n= 2*n*1/2 -n = n-n =0

the variance will be

σ²(Xn) = ∑ [Xn - E(Xn)]² p(Xn) = ∑ [(2X1-n) - 0 ]² p(X1) = ∑ (4*X1²-4*X1*n+n²) p(X1) = = 4*∑ X1²p(X1) - 4n ∑X1 p(X1) -  n²∑p(X1) = 2*E(X1²) -4n*E(X1)- n²

since

σ²(X1) = n*p*(1-p) = n*0,5*0,5=n/4

and

σ²(X1) = E(X1²) - [E(X1)]²

n/4 = E(X1²) - (n/2)²

E(X1²) = n(n+1)/4

therefore

σ²(Xn) = 4*E(X1²) -4n*E(X1)- n² = 4*n(n+1)/4 - 4*n*n/2 - n² = n(n+1) - 2n² - n²

= n - 2n² = n(1-2n)

σ²(Xn) = n(1-2n)

4 0
1 year ago
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