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2 years ago

What is the rate of change in y per unit change in x for the function 30x+5y=15

Mathematics

1 answer:

ad-work [718]2 years ago

6 0

Step-by-step explanation:

To find the answer you have to put it into y=mx+b format.

This means that it will be 5y=-30x+15

So it will simplify to y=-6x+3

This means that:

If y=2

You substitute then…

2=-6x+3

So x would equal 1/6.

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What is the common difference between successive terms in the sequence? 0.36, 0.26, 0.16, 0.06, –0.04, –0.14,

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2 years ago

A volleyball player sets the ball in the air, and the height of the ball after t seconds is given in feet by h= -16^2+12t+6. A t

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Answer:

Step-by-step explanation:

The given function is

$h=-16t^2+12t+6$

The graph of this function is a parabola that opens downwards

The line $h(t)=8$ intersects this parabola, when

$t=0.5,t=0.25$

The teammate can spike the ball after 0.25 seconds or 0.5 seconds.

The two solutions are reasonable. When the volleyball is accelerating into the air, it passes a height of 8 after 0.25 seconds.

When the ball is dropping after it attains maximum height, it attains another height of 8 after 0.5 seconds again.

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2 years ago

One brand of cookies is sold in three different box sizes: small, medium, and large. The small box has half the volume of the me

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1 year ago

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).

For X=0

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

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

For X=1

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:

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

For X=2

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.

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

For X=3

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.

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

For X=4

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.

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

For X=5

This happens when

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

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

For X=6

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

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

For X=7

This happens when

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

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

For X=8

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

2 years ago