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Vsevolod [243]
3 years ago
9

A zoo keeper published the following stem-and-leaf plot showing the number of lizard at each major zoo in the country. How many

zoos have exactly 42 lizards

Mathematics
1 answer:
Alina [70]3 years ago
3 0

This question is incomplete because it lacks the diagram showing the leaf and stem plots

Find attached to this answer the diagram showing the Stem and leaf plots

Answer:

Only one Zoo has exactly 42 lizards

Step-by-step explanation:

If you check out the attached diagram , you will observe that there is a separation using a vertical line

In the diagram also we have been provided with a key which was explained as :

2|0 = 20 lizards

On the left hand side, we have number ranging from 0 - 6 arranged in a vertical form. On the right hand side we almost have numbers that are arranged in front of each number horizontally

Let me further explain the attached diagram

It is important to note that the numbers arranged vertically on the left are in the units of tens. Hence, using the given key :2l0

0|

1|

2| 0 6 8 8 8

This means , we have 20 lizards in one zoo, 26 lizards in one zoo, 28 lizards in 3 zoos

3| 0 2 6 6 7 8

30 lizards in one zoo, 32 lizards in one zoo, 36 lizards in 2 zoos , 37 lizards in one zoo and 38 lizards in one zoo

4| 1 2 6 6

We have 41 lizards in one zoo, 42 lizards in one zoo, and 46 lizards in two zoos

5l

6|

Based on the explanation given above, we can see that only one Zoo has exactly 42 lizards

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8÷2 =4

4-6 = -2

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Step-by-step explanation:

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A bin contains 25 light bulbs, 5 of which are in good condition and will function for at least 30 days, 10 of which are partiall
Ira Lisetskai [31]

Answer:

The probability that it will still be working after one week is \frac{1}{5}

Step-by-step explanation:

Given :

Total number of bulbs = 25

Number of bulbs which are good condition and will function for at least 30 days = 5

Number of bulbs which are partially defective and will fail in their second day of use = 10

Number of bulbs which are totally defective and will not light up = 10

To find : What is the probability that it will still be working after one week?

Solution :

First condition is a randomly chosen bulb initially lights,

i.e. Either it is in good condition and partially defective.

Second condition is it will still be working after one week,

i.e. Bulbs which are good condition and will function for at least 30 days

So, favorable outcome is 5

The probability that it will still be working after one week is given by,

\text{Probability}=\frac{\text{Favorable outcome}}{\text{Total number of outcome}}

\text{Probability}=\frac{5}{25}

\text{Probability}=\frac{1}{5}

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Return to the credit card scenario of Exercise 12 (Section 2.2), and let C be the event that the selected student has an America
Nadya [2.5K]

Answer:

A. P = 0.73

B. P(A∩B∩C') = 0.22

C. P(B/A) = 0.5

   P(A/B) = 0.75

D. P(A∩B/C) = 0.4

E. P(A∪B/C) = 0.85

Step-by-step explanation:

Let's call A the event that a student has a Visa card, B the event that a student has a MasterCard and C the event that a student has a American Express card. Additionally, let's call A' the event that a student hasn't a Visa card, B' the event that a student hasn't a MasterCard and C the event that a student hasn't a American Express card.

Then, with the given probabilities we can find the following probabilities:

P(A∩B∩C') = P(A∩B) - P(A∩B∩C) = 0.3 - 0.08 = 0.22

Where P(A∩B∩C') is the probability that a student has a Visa card and a Master Card but doesn't have a American Express, P(A∩B) is the probability that a student has a has a Visa card and a MasterCard and P(A∩B∩C) is the probability that a student has a Visa card, a MasterCard and a American Express card. At the same way, we can find:

P(A∩C∩B') = P(A∩C) - P(A∩B∩C) = 0.15 - 0.08 = 0.07

P(B∩C∩A') = P(B∩C) - P(A∩B∩C) = 0.1 - 0.08 = 0.02

P(A∩B'∩C') = P(A) - P(A∩B∩C') - P(A∩C∩B') - P(A∩B∩C)

                   = 0.6 - 0.22 - 0.07 - 0.08 = 0.23

P(B∩A'∩C') = P(B) - P(A∩B∩C') - P(B∩C∩A') - P(A∩B∩C)

                   = 0.4 - 0.22 - 0.02 - 0.08 = 0.08

P(C∩A'∩A') = P(C) - P(A∩C∩B') - P(B∩C∩A') - P(A∩B∩C)

                   = 0.2 - 0.07 - 0.02 - 0.08 = 0.03

A. the probability that the selected student has at least one of the three types of cards is calculated as:

P = P(A∩B∩C) + P(A∩B∩C') + P(A∩C∩B') + P(B∩C∩A') + P(A∩B'∩C') +              

     P(B∩A'∩C') + P(C∩A'∩A')

P = 0.08 + 0.22 + 0.07 + 0.02 + 0.23 + 0.08 + 0.03 = 0.73

B. The probability that the selected student has both a Visa card and a MasterCard but not an American Express card can be written as P(A∩B∩C') and it is equal to 0.22

C. P(B/A) is the probability that a student has a MasterCard given that he has a Visa Card. it is calculated as:

P(B/A) = P(A∩B)/P(A)

So, replacing values, we get:

P(B/A) = 0.3/0.6 = 0.5

At the same way, P(A/B) is the probability that a  student has a Visa Card given that he has a MasterCard. it is calculated as:

P(A/B) = P(A∩B)/P(B) = 0.3/0.4 = 0.75

D. If a selected student has an American Express card, the probability that she or he also has both a Visa card and a MasterCard is  written as P(A∩B/C), so it is calculated as:

P(A∩B/C) = P(A∩B∩C)/P(C) = 0.08/0.2 = 0.4

E. If a the selected student has an American Express card, the probability that she or he has at least one of the other two types of cards is written as P(A∪B/C) and it is calculated as:

P(A∪B/C) = P(A∪B∩C)/P(C)

Where P(A∪B∩C) = P(A∩B∩C)+P(B∩C∩A')+P(A∩C∩B')

So, P(A∪B∩C) = 0.08 + 0.07 + 0.02 = 0.17

Finally, P(A∪B/C) is:

P(A∪B/C) = 0.17/0.2 =0.85

4 0
3 years ago
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