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Elis [28]
2 years ago
6

A school district has 2 teaching positions to fill and there are 8 applicants to choose from. How many different possibilities a

re there?
A. 4 (correct answer right?)
B. 28
C. 3
D. 56
Mathematics
2 answers:
Olenka [21]2 years ago
6 0

I think the answer would be 56. Bc the first job has 8 different people who could get the job and then once someone is chosen for that slot the next teaching position only has 7 candidates. So you would take 8*7 which equals 56.

IgorLugansk [536]2 years ago
5 0

Answer:

The correct answer is D

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If cos(t) = 2/7 and t is in the 4th quadrant, find sin(t).
Yuliya22 [10]
We can use the Pythagorean Trigonometric Identity which says:
sin^2(t)+cos^2(t)=1

Since we need to find sin(t), we have to solve for it:
sin(t)= \sqrt{1-cos^2(t)}

Let's plug in the given cos(t) value:
sin(t) = \sqrt{1-cos^2( \frac{2}{7})}

And solve sin(t):
sin(t) = \sqrt{1- \frac{4}{49} } = \frac{x}{y} \sqrt{ \frac{49}{49}- \frac{4}{49} }

Simplify further:
sin(t) = \sqrt{ \frac{45}{49} } = \frac{ \sqrt{45} }{7} = \frac{ \sqrt{9*5} }{7}

And it all simplifies down to:
sin(t) = \frac{3 \sqrt{5} }{7}

Since it's in the 4th quadrant, the sin(t) value is going to be negative. So, your final answer is: 
sin(t) = - \frac{ 3\sqrt{5} }{7}

Hope this helps!
7 0
2 years ago
susan took two tests.the probability of her passing both tests is 0.6.the probability of her passing the first test is 0.8.what
OverLord2011 [107]
Formula for this is as follows:
probability of her passing both 0.6/0.8 - first test and this is a fraction. 0.6/0.8
0.6/0.8= divide 0.6 by 0.8=0.75
that means probability of her passing the second test is 75%
6 0
1 year ago
At a local fitness center, members pay a $14 membership fee and $5 for each aerobics class. Nonmembers pay $6 for each aerobics
nata0808 [166]

Answer:

that would be $25 for each.

5 0
1 year ago
Read 2 more answers
Three couples and two single individuals have been invited to an investment seminar and have agreed to attend. Suppose the proba
Maksim231197 [3]

Answer:

(a) Probability mass function

P(X=0) = 0.0602

P(X=1) = 0.0908

P(X=2) = 0.1704

P(X=3) = 0.2055

P(X=4) = 0.1285

P(X=5) = 0.1550

P(X=6) = 0.1427

P(X=7) = 0.0390

P(X=8) = 0.0147

NOTE: the sum of the probabilities gives 1.0068 for rounding errors. It can be divided by 1.0068 to get the adjusted values.

(b) Cumulative distribution function of X

F(X=0) = 0.0602

F(X=1) = 0.1510

F(X=2) = 0.3214

F(X=3) = 0.5269

F(X=4) = 0.6554

F(X=5) = 0.8104

F(X=6) = 0.9531

F(X=7) = 0.9921

F(X=8) = 1.0068

Step-by-step explanation:

Let X be the number of people who arrive late to the seminar, we can assess that X can take values from 0 (everybody on time) to 8 (everybody late).

<u>For X=0</u>

This happens when every couple and the singles are on time (ot).

P(X=0)=P(\#1=ot)*P(\#2=ot)*P(\#3=ot)*P(\#4=ot)*P(\#5=ot)\\\\P(X=0)=(1-0.43)^{5}=0.57^5= 0.0602

<u>For X=1</u>

This happens when only one single arrives late. It can be #4 or #5. As the probabilities are the same (P(#4=late)=P(#5=late)), we can multiply by 2 the former probability:

P(X=1) = P(\#4=late)+P(\#5=late)=2*P(\#4=late)\\\\P(X=1) = 2*P(\#1=ot)*P(\#2=ot)*P(\#3=ot)*P(\#4=late)*P(\#5=ot)\\\\P(X=1) = 2*0.57*0.57*0.57*0.43*0.57\\\\P(X=1) = 2*0.57^4*0.43=2*0.0454=0.0908

<u>For X=2</u>

This happens when

1) Only one of the three couples is late, and the others cooples and singles are on time.

2) When both singles are late , and the couples are on time.

P(X=2)=3*(P(\#1=l)*P(\#2=ot)*P(\#3=ot)*P(\#4=ot)*P(\#5=ot))+P(\#1=ot)*P(\#2=ot)*P(\#3=ot)*P(\#4=l)*P(\#5=l)\\\\P(X=2)=3*(0.43*0.57^4)+(0.43^2*0.57^3)=0.1362+0.0342=0.1704

<u>For X=3</u>

This happens when

1) Only one couple (3 posibilities) and one single are late (2 posibilities). This means there are 3*2=6 combinations of this.

P(X=3)=6*(P(\#1=l)*P(\#2=ot)*P(\#3=ot)*P(\#4=l)*P(\#5=ot))\\\\P(X=3)=6*(0.43^2*0.57^3)=6*0.342=0.2055

<u>For X=4</u>

This happens when

1) Only two couples are late. There are 3 combinations of these.

2) Only one couple and both singles are late. Only one combination of these situation.

P(X=4)=3*(P(\#1=l)*P(\#2=l)*P(\#3=ot)*P(\#4=ot)*P(\#5=ot))+P(\#1=l)*P(\#2=ot)*P(\#3=ot)*P(\#4=l)*P(\#5=l)\\\\P(X=4)=3*(0.43^2*0.57^3)+(0.43^3*0.57^2)\\\\P(X=4)=3*0.0342+ 0.0258=0.1027+0.0258=0.1285

<u>For X=5</u>

This happens when

1) Only two couples (3 combinations) and one single are late (2 combinations). There are 6 combinations.

P(X=6)=6*(P(\#1=l)*P(\#2=l)*P(\#3=ot)*P(\#4=l)*P(\#5=ot))\\\\P(X=6)=6*(0.43^3*0.57^2)=6*0.0258=0.1550

<u>For X=6</u>

This happens when

1) Only the three couples are late (1 combination)

2) Only two couples (3 combinations) and one single (2 combinations) are late

P(X=6)=P(\#1=l)*P(\#2=l)*P(\#3=l)*P(\#4=ot)*P(\#5=ot)+6*(P(\#1=l)*P(\#2=l)*P(\#3=ot)*P(\#4=l)*P(\#5=ot))\\\\P(X=6)=(0.43^3*0.57^2)+6*(0.43^4*0.57)\\\\P(X=6)=0.0258+6*0.0195=0.0258+0.1169=0.1427

<u>For X=7</u>

This happens when

1) Only one of the singles is on time (2 combinations)

P(X=7)=2*P(\#1=l)*P(\#2=l)*P(\#3=l)*P(\#4=l)*P(\#5=ot)\\\\P(X=7)=2*0.43^4*0.57=0.0390

<u>For X=8</u>

This happens when everybody is late

P(X=8)=P(\#1=l)*P(\#2=l)*P(\#3=l)*P(\#4=l)*P(\#5=l)\\\\P(X=8) = 0.43^5=0.0147

8 0
1 year ago
An ice cream store has two new flavors: Fantasy and Ecstasy. Each barrel of Fantasy requires 4
Y_Kistochka [10]

Answer:

  • Fantasy: 3 barrels
  • Ecstasy: 1 barrel

Step-by-step explanation:

<u>Given</u>

  Fantasy uses 4 lb of nuts, 3 lb of chocolate, for a profit of $50

  Ecstasy uses 4 lb of nuts, 1 lb of chocolate, for a profit of $40

  In stock are 16 lb of nuts, 10 lb of chocolate

<u>Find</u>

 amount of each to maximize profit

<u>Solution</u>

Let x and y represent barrels of Fantasy and Ecstasy, respectively. Then the limitations on production are ...

  4x +4y ≤ 16 . . . lb of nuts

  3x +y ≤ 10 . . . . lb of chocolate

We want to maximize

  50x +40y

The graph shows the feasible region. Its vertices are ...

  (0, 4), (3, 1), (3.33, 0)

Profit is maximized at $190 when production is 3 barrels of Fantasy and 1 barrel of Ecstasy.

8 0
2 years ago
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