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elena-14-01-66 [18.8K]
3 years ago
14

A chef boiled 4 eggs and put them in a basket with 8 eggs that were not boiled. All of the eggs look the same. Randy selects an

egg, keeps it, and then selects another egg. Which expression gives the probability that he selects 2 boiled eggs?
A. 4/12*3/11
B. 4/12*4/12
C.4/12*3/12
D.4/12*4/11
Mathematics
2 answers:
tamaranim1 [39]3 years ago
7 0

Answer:

i took the test and its A

Step-by-step explanation:

All together there are 12 eggs total with 4 that are non boiled and 8 that are boiled. If he takes out 2 eggs, it then goes from 12 total eggs to 11 then take 1 away from 4 to have 3 so its A. 4/12 * 3/11

oee [108]3 years ago
5 0
Expression A gives the correct probability.

On the first draw, there is a 4/12 chance of choosing a boiled egg because there are 4 boiled eggs and a total of 12 eggs to choose from.

On the second draw, there is a 3/11 chance of choosing a boiled egg because there are 3 boiled eggs and a total of 11 eggs to choose from.
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20. The total mass of three boxes, A, B, and C, is 1.95 kg. The mass of Box A is 700 g. The mass of Box B is 4 times the mass of
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Answer:

<h3>Mass of B in Kg = -558.44kg</h3>

Step-by-step explanation:

LET'S DO THIS!

Total mass of A, B and C = 1.95kg

Mass of A = 700 kg

Mass of B = 4x the mass of c (4 x C) which is 4c

Mass of C = ? ( let's call it C )

<h3>Adding all together </h3>

700 + 4c + C = 1.95

<h3>Add like terms</h3>

700 + 5C = 1.95

5C = 1.95 - 700

5C = -698.05

C = -698.05 ÷ 5

<h3>C = -139.61</h3>

<h3>To find B now </h3><h3>Remember they said B is 4 times the mass of C and C = -139.61</h3>

therefore B = 4 × -139.61

<h3>B = -558.44 kg </h3>

<h3>To check if we are correct, we add the masses of A, B and C to see if it equals their total mass which is 1.95kg</h3>

<h3>Using your calculator: </h3>

= 700 + ( -558.44 ) + ( -139.61 )

= 700 - 558.44 -139.61

= 1.95 kg

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2 years ago
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An instructor who taught two sections of engineering statistics last term, the first with 25 students and the second with 35, de
Amiraneli [1.4K]

Answer:

a) P=0.1721

b) P=0.3528

c) P=0.3981

Step-by-step explanation:

This sampling can be modeled by a binominal distribution where p is the probability of a project to belong to the first section and q the probability of belonging to the second section.

a) In this case we have a sample size of n=15.

The value of p is p=25/(25+35)=0.4167 and q=1-0.4167=0.5833.

The probability of having exactly 10 projects for the second section is equal to having exactly 5 projects of the first section.

This probability can be calculated as:

P=\frac{n!}{(n-k)!k!}p^kq^{n-k}= \frac{15!}{(10)!5!}\cdot 0.4167^5\cdot0.5833^{10}=0.1721

b) To have at least 10 projects from the 2nd section, means we have at most 5 projects for the first section. In this case, we have to calculate the probability for k=0 (every project belongs to the 2nd section), k=1, k=2, k=3, k=4 and k=5.

We apply the same formula but as a sum:

P(k\leq5)=\sum_{k=0}^{5}\frac{n!}{(n-k)!k!}p^kq^{n-k}

Then we have:

P(k=0)=0.0003\\P(k=1)=0.0033\\P(k=2)=0.0165\\P(k=3)=0.0511\\P(k=4)=0.1095\\P(k=5)=0.1721\\\\P(k\leq5)=0.0003+0.0033+0.0165+0.0511+0.1095+0.1721=0.3528

c) In this case, we have the sum of the probability that k is equal or less than 5, and the probability tha k is 10 or more (10 or more projects belonging to the 1st section).

The first (k less or equal to 5) is already calculated.

We have to calculate for k equal to 10 or more.

P(k\geq10)=\sum_{k=10}^{15}\frac{n!}{(n-k)!k!}p^kq^{n-k}

Then we have

P(k=10)=0.0320\\P(k=11)=0.0104\\P(k=12)=0.0025\\P(k=13)=0.0004\\P(k=14)=0.0000\\P(k=15)=0.0000\\\\P(k\geq10)=0.032+0.0104+0.0025+0.0004+0+0=0.0453

The sum of the probabilities is

P(k\leq5)+P(k\geq10)=0.3528+0.0453=0.3981

8 0
3 years ago
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