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Alchen [17]
3 years ago
11

. A fair coin and then a die with 6 sides are tosses find the probabilities of the six events occurring respectively a. P(Tails)

b. P(3) c. P(tails and 3) d. P(tails | 3) e. P(3 | tails) f. P(4 or a 5)
Mathematics
1 answer:
Dvinal [7]3 years ago
7 0

Answer:

F

Step-by-step explanation:

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each school bus going on the field trip hold 36 students and 4 adults. There are 6 filled buses on the field trip. How many peop
Liono4ka [1.6K]
Since one bus holds 36 students and 4 adults, 40 people are in one bus so...
40 (6)
=240

Total 240 people are going on the field trip.

Hope this helped you.
Have a great day!

3 0
3 years ago
Read 2 more answers
Suppose a particular type of cancer has a 0.9% incidence rate. Let D be the event that a person has this type of cancer, therefo
natita [175]

Answer:

There is a 12.13% probability that the person actually does have cancer.

Step-by-step explanation:

We have these following probabilities.

A 0.9% probability of a person having cancer

A 99.1% probability of a person not having cancer.

If a person has cancer, she has a 91% probability of being diagnosticated.

If a person does not have cancer, she has a 6% probability of being diagnosticated.

The question can be formulated as the following problem:

What is the probability of B happening, knowing that A has happened.

It can be calculated by the following formula

P = \frac{P(B).P(A/B)}{P(A)}

Where P(B) is the probability of B happening, P(A/B) is the probability of A happening knowing that B happened and P(A) is the probability of A happening.

In this problem we have the following question

What is the probability that the person has cancer, given that she was diagnosticated?

So

P(B) is the probability of the person having cancer, so P(B) = 0.009

P(A/B) is the probability that the person being diagnosticated, given that she has cancer. So P(A/B) = 0.91

P(A) is the probability of the person being diagnosticated. If she has cancer, there is a 91% probability that she was diagnosticard. There is also a 6% probability of a person without cancer being diagnosticated. So

P(A) = 0.009*0.91 + 0.06*0.991 = 0.06765

What is the probability that the person actually does have cancer?

P = \frac{P(B).P(A/B)}{P(A)} = \frac{0.91*0.009}{0.0675} = 0.1213

There is a 12.13% probability that the person actually does have cancer.

3 0
3 years ago
Can someone please help me??
Nesterboy [21]

Answer:

Answers are below

Step-by-step explanation:

x     y

0    $25,001.00

1     $25,000.95

2    $25,000.90

3    $25,000.86

I graphed the function on the graph below and found the values of y for each value of x.

*My graph showed that the initial value of the printer was $25,001.00, instead of $25,000.00*

7 0
3 years ago
Connor drove 340 miles on 8 gallons of gas. How many miles did his car travel per gallon of gas?
MrMuchimi

Answer:

42.5

Step-by-step explanation:

calculatour

5 0
2 years ago
Classify the following triangle. Check all that apply.
klasskru [66]

Answer:

Step-by-step explanation:

a. scalene  :  Reason: all angles and sides are different measurements.

f. acute : all the three angles are acute (Less than 90)

5 0
3 years ago
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