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Afina-wow [57]
3 years ago
11

A six-sided fair die and an eight-sided fair die are rolled together. What is the probability of getting numbers whose sum is a

multiple of 3?
Mathematics
2 answers:
aalyn [17]3 years ago
3 0
There are 48 possible outcomes in this situation, and of those outcomes, the ones whose sums are a multiple of three are;
12, 21, 15, 51, 24, 42, 33, 18, 27, 36, 63, 45, 54, 48, 57, and 66. So, that is 16 out of 48 possibilities, or 16/48, which simplifies to 1/3. Written as a percent, the probability of getting numbers whose sum is a multiple of three is 33.33%.

Hope this is helpful! :)
ohaa [14]3 years ago
3 0

Answer:

1/3

Step-by-step explanation:

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liraira [26]

Answer:

D

Step-by-step explanation:

If y = log x is the basic function, let's see the transformation rule(s):

Then,

1. y = log (x-a) is the original shifted a units to the right.

2. y = log x + b is the original shifted b units up

Hence, from the equation, we can say that this graph is:

** 2 units shifted right (with respect to original), and

** 10 units shifted up (with respect to original)

<u><em>only, left or right shift affects vertical asymptotes.</em></u>

Since, the graph of y = log x has x = 0 as the vertical asymptote and the transformed graph is shifted 2 units right (to x = 2), x = 2 is the new vertical asymptote.

Answer choice D is right.

7 0
3 years ago
2.5% of a population are infected with a certain disease. There is a test for the disease, however the test is not completely ac
Aleks04 [339]

Answer:

P(disease/positivetest) = 0.36116

Step-by-step explanation:

This is a conditional probability exercise.

Let's name the events :

I : ''A person is infected''

NI : ''A person is not infected''

PT : ''The test is positive''

NT : ''The test is negative''

The conditional probability equation is :

Given two events A and B :

P(A/B) = P(A ∩ B) / P(B)

P(B) >0

P(A/B) is the probability of the event A given that the event B happened

P(A ∩ B) is the probability of the event (A ∩ B)

(A ∩ B) is the event where A and B happened at the same time

In the exercise :

P(I)=0.025

P(NI)= 1-P(I)=1-0.025=0.975\\P(NI)=0.975

P(PT/I)=0.904\\P(PT/NI)=0.041

We are looking for P(I/PT) :

P(I/PT)=P(I∩ PT)/ P(PT)

P(PT/I)=0.904

P(PT/I)=P(PT∩ I)/P(I)

0.904=P(PT∩ I)/0.025

P(PT∩ I)=0.904 x 0.025

P(PT∩ I) = 0.0226

P(PT/NI)=0.041

P(PT/NI)=P(PT∩ NI)/P(NI)

0.041=P(PT∩ NI)/0.975

P(PT∩ NI) = 0.041 x 0.975

P(PT∩ NI) = 0.039975

P(PT) = P(PT∩ I)+P(PT∩ NI)

P(PT)= 0.0226 + 0.039975

P(PT) = 0.062575

P(I/PT) = P(PT∩I)/P(PT)

P(I/PT)=\frac{0.0226}{0.062575} \\P(I/PT)=0.36116

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i read in class 4 but I don't explain the answer

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andriy [413]

Answer:

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Step-by-step explanation:

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ELEN [110]

Answer:

Im 100% sure that it’s (A)

5 0
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