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stealth61 [152]
3 years ago
11

Consider the experiment of rolling two standard (six-sided) dice and taking their sum. Assume that each die lands on each of its

faces equally often. We consider the outcomes of this experiment to be the ordered pairs of numbers on the dice, and the events of interest to be the different sums. (a) How many outcomes does this experiment have
Mathematics
1 answer:
Strike441 [17]3 years ago
7 0

Answer:

36

Step-by-step explanation:

Given that:

2 standard six sided dice are rolled :

Die 1 = 1, 2, 3, 4, 5, 6

Die 2 = 1, 2, 3, 4, 5, 6

Outcome = ordered pair of Number on the two dice

Event = different sum of the ordered pair of numbers.

The number of outcomes in the experiment = the sample space =) number of Faces)^number of dies

= 6^2

= 36 different outcomes

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If the width of a rectangle is 3 yards less than it’s length and the perimeter is 130 yards, what is the length and the width.
Marianna [84]

Answer: P=130

2W +2L=130

2(31) + 2(34)=130

W=31

L=34

Step-by-step explanation:

P=130

2L + 2W

W=L-3

Add all of the sides together to get perimeter.

2(L-3) + 2L=130

2L-6+2L=130

4L-6+6=130+6

4L=136

4L/4=136/4

L=34

W=L-3

W=34-3

W=31



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3 years ago
Trigonometry find the missing side​
Katarina [22]

Step-by-step explanation:

griminology is one of part of body

5 0
3 years ago
Una caja en forma de prisma rectangular tiene 10cm3 de volumen.Si la longitud de cada arista se multiplica por 4, ¿cual sera el
Salsk061 [2.6K]
40cm. Por que 10 e 4
4 0
3 years ago
HELP ME NOW PLEASE, IM BEGGING.
svlad2 [7]
1. $1.23 per pair
A dozen is 12 so you do 12x7=84

Then you do 103.32 divided by 84 which gives you 1.23

2. His error was doing 103.32 divided by 7 because he thought it was only 7 pairs but it’s 7 dozen.
3 0
3 years ago
Read 2 more answers
The time taken to assemble a laptop computer in a certain plant is a random variable having a normal distribution of 20 hours an
ludmilkaskok [199]

Answer:

a) 40.13% probability that a laptop computer can be assembled at this plant in a period of time of less than 19.5 hours.

b) 34.13% probability that a laptop computer can be assembled at this plant in a period of time between 20 hours and 22 hours.

Step-by-step explanation:

When the distribution is normal, we use the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this question:

\mu = 20, \sigma = 2

a)Less than 19.5 hours?

This is the pvalue of Z when X = 19.5. So

Z = \frac{X - \mu}{\sigma}

Z = \frac{19.5 - 20}{2}

Z = -0.25

Z = -0.25 has a pvalue of 0.4013.

40.13% probability that a laptop computer can be assembled at this plant in a period of time of less than 19.5 hours.

b)Between 20 hours and 22 hours?

This is the pvalue of Z when X = 22 subtracted by the pvalue of Z when X = 20. So

X = 22

Z = \frac{X - \mu}{\sigma}

Z = \frac{22 - 20}{2}

Z = 1

Z = 1 has a pvalue of 0.8413

X = 20

Z = \frac{X - \mu}{\sigma}

Z = \frac{20 - 20}{2}

Z = 0

Z = 0 has a pvalue of 0.5

0.8413 - 0.5 = 0.3413

34.13% probability that a laptop computer can be assembled at this plant in a period of time between 20 hours and 22 hours.

4 0
3 years ago
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