Can anyone help me with this?

Mathematics

1 answer:

rewona [7]3 years ago

8 0

Answer:u

hard

Step-by-step explanation:

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How do I solve this?

algol [13]

I think the answer is (C)

4 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

4 years ago

Read 2 more answers

In a class of P students, the average of test scores is 70. In another class the average test scores is 92. When scores of the t

Basile [38]

So, since averages are defined as:

$\frac{\sum_{k=1}^{P} P_k}{P}=70$

So, since P are the total number of elements and P_k is the P_kth student. This is saying if we sum over each student's score and divide it by the number of students, we should get P, which is true.

So, using that logic, the other class can be represented as:

$\frac{\sum_{k=1}^{N} N_k}{N}=70$