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2 years ago

2. Which situation does have a final value of 0?

Mathematics

1 answer:

professor190 [17]2 years ago

7 0

Answer:

Step-by-step explanation:

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paloma eats 2/8 of an apple in the morning and 3/8 more in the afternoon. how much of an apple did she eat in total

marishachu [46]

Step-by-step explanation:

in this problem, we are solving for the amount of apple that Paloma eats

1. she ate 2/8 of a complete apple -hence the will be a remainder, we will find

that remainder

2. she also ate 3/8 of the remainder for lunch

3. we then proceed to calculate all apple she ate in total

$1. \frac{2}{8} *1= \frac{1}{4} \\\\1-\frac{1}{4}= \frac{4-1}{4} = \frac{3}{4}$

2 she ate 3/8 of the remaining 3/4 apple

$1. \frac{3}{8} *\frac{3}{4} = \frac{9}{32}$

3. in total she ate 1/4+ 9/32

$\frac{1}{4} +\frac{9}{32} = \frac{8+9}{32} = \frac{17}{32}$

6 0

2 years ago

A new factory is growing fast and hiring a lot of new workers. Every week, the total number of workers doubles in size. If it ta

MAXImum [283]

We know that the number of workers doubles every week.

This implies that the number of workers of the previous week is exactly half of the current one's.

So, it took 11 weeks for the factory to be at half capacity.

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2 years ago

A. identify the parent function of the given graph.

Paha777 [63]

The parent function is f(x) = x^3

The domain are all x values (-infinity, infinity)

The range are all y values (-infinity, infinity)

4 0

2 years ago

Read 2 more answers

The weights of certain machine components are normally distributed with a mean of 8.01 g and a standard deviation of 0.06 g. Fin

Free_Kalibri [48]

Answer:

Option D) 7.90 g and 8.12 g

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 8.01 g

Standard Deviation, σ = 0.06 g

We are given that the distribution of weights is a bell shaped distribution that is a normal distribution.

Formula:

$z_{score} = \displaystyle\frac{x-\mu}{\sigma}$

We have to find the value of x such that the probability is 0.03

P(X > x)

$P( X > x) = P( z > \displaystyle\frac{x - 8.01}{0.06})=0.03$

$= 1 -P( z \leq \displaystyle\frac{x - 8.01}{0.06})=0.03$

$=P( z \leq \displaystyle\frac{x - 8.01}{0.06})=0.97$

Calculation the value from standard normal z table, we have,

$\displaystyle\frac{x - 8.01}{0.06} = 1.881\\\\x = 8.12$

Thus, 8.17 g separates the top 3% of the weights.

P(X < x)

$P( X < x) = P( z < \displaystyle\frac{x - 8.01}{0.06})=0.03$

Calculation the value from standard normal z table, we have,

$\displaystyle\frac{x - 8.01}{0.06} = -1.881\\\\x = 7.90$

Thus, 7.90 separates the bottom 3% of the weights.

Thus, the correct answer is

Option D) 7.90 g and 8.12 g

7 0

2 years 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