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Tamiku [17]
2 years ago
7

A bag of fruit contains 4 apples, 1 plum, 2 apricots, and 3 oranges. Pieces of fruit are drawn twice with replacement. What is P

(apple, then apricot) A.4/5 B.2/25 C.3/25 D.3/5
Mathematics
1 answer:
bezimeni [28]2 years ago
6 0
Since there is replacement, drawings are independent which means that the result of one of them does not affect the others.
Probability of drawing an apple is 4/10 (4 apple out of 10 fruits)
Probability of drawing an apricot is 2/10
Probability on one happening consecutively to the other (thanks to independence) is 4/10*2/10=8/100
Which can be simplified to 2/25
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25. You have to do 65x65-(60x60) and the square root it
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Please help me with the following questions. Thanks!
Natali5045456 [20]

Answer:

The answer to your questions  are in bold

Step-by-step explanation:

a)  

 C = \left[\begin{array}{ccc}-6&6\\-2&4\\\end{array}\right]

    = -24 + 12

   = -12

b)  -1   7          -4   -1                   -1  - 4        7 - 1               - 5      6

    -2 -6          -8    8      =          -2 - 8       -6 + 8              -10      2

     2 -3           2   -7                  2 + 2        -3 - 7               4       -10

   -1   10         -6     5                 -1 - 6         10 + 5              -7       15

5 0
3 years ago
jay beats terry at table tennis 3 out of 5 games. what is the experimental probability that terry will win their next game
lozanna [386]

Answer:

The experimental probability that terry will win their next game is \frac{2}{5}

Step-by-step explanation:

While theoretical probability is the ratio of the number of favourable outcomes to the total number of outcomes, the experimental probability is the ratio of number of times the event is occurring to the total number of trials of the experiment.

Out of 5 games jay wins  3

Now the experimental  probability that terry will win their next game

 = 1 - Probability of Jay winning the game----------------------(1)

Probability of Jay winning the game = \frac{\text{number of occurrences of the event}}{\text{total number of trials made}}

=> \frac{3}{5}----------------------------------------------------(2)

Substituting (2) in (1)

=> 1 -  \frac{3}{5}

=>  \frac{5-3}{5}

=>  \frac{2}{5}

6 0
3 years ago
Eric throws a biased coin 10 times. He gets 3 tails. Sue throw the same coin 50 times. She gets 20 tails. Aadi is going to throw
iren2701 [21]

Answer:

(1) The correct option is (A).

(2) The probability that Aadi will get Tails is \frac{2}{5}.

Step-by-step explanation:

It is provided that:

  • Eric throws a biased coin 10 times. He gets 3 tails.
  • Sue throw the same coin 50 times. She gets 20 tails.

The probability of tail in both cases is:

  • P(\text{Tail})=\frac{3}{10}
  • P(\text{Tail})=\frac{20}{50}=\frac{2}{5}

(1)

According to the Central limit theorem, if from an unknown population large samples of sizes n > 30, are selected and the sample proportion for each sample is computed then the sampling distribution of sample proportion follows a Normal distribution.

In this case we need to compute the proportion of tails.

Then according to the Central limit theorem, Sue's estimate is best because she throws it <em>n = </em>50 > 30 times.

Thus, the correct option is (A).

(2)

As explained in the first part that Sue's estimate is best for getting a tail, the probability that Aadi will get Tails when he tosses the coin once is:

P(\text{Tail})=\frac{20}{50}=\frac{2}{5}

Thus, the probability that Aadi will get Tails is \frac{2}{5}.

7 0
3 years ago
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