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Tju [1.3M]
3 years ago
15

The sum of 7 times a number and 3 equals 8

Mathematics
1 answer:
mars1129 [50]3 years ago
6 0

Answer:

7x+3=8

Step-by-step explanation:

this is the answer

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Find the coordinates of the image under the transformation (x – 4, y + 6) for ∆XYZ at X(-6, 1), Y(4, 0), Z(1, 3)
Afina-wow [57]

Answer:

I think b

Step-by-step explanation:

I prefer B but it also might be D

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Sasha invests $200 at 5% after 6 months
Umnica [9.8K]
5% of 200 is 10, after six months, so total of $60? not sure though.

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Suppose that a basketball player can score on a particular shot with probability .3. Use the central limit theorem to find the a
Rom4ik [11]

Answer:

(a) The probability that the number of successes is at most 5 is 0.1379.

(b) The probability that the number of successes is at most 5 is 0.1379.

(c) The probability that the number of successes is at most 5 is 0.1379.

(d) The probability that the number of successes is at most 11 is 0.9357.

→ All the exact probabilities are more than the approximated probability.

Step-by-step explanation:

Let <em>S</em> = a basketball player scores a shot.

The probability that a basketball player scores a shot is, P (S) = <em>p</em> = 0.30.

The number of sample selected is, <em>n</em> = 25.

The random variable S\sim Bin(25,0.30)

According to the central limit theorem if the sample taken from an unknown population is large then the sampling distribution of the sample proportion (\hat p) follows a normal distribution.

The mean of the the sampling distribution of the sample proportion is: E(\hat p)=p=0.30

The standard deviation of the the sampling distribution of the sample proportion is:

SD(\hat p)=\sqrt{\frac{ p(1- p)}{n} }=\sqrt{\frac{ 0.30(1-0.30)}{25} }=0.092

(a)

Compute the probability that the number of successes is at most 5 as follows:

The probability of 5 successes is: p=\frac{5}{25} =0.20

P(\hat p\leq 0.20)=P(\frac{\hat p-E(\hat p)}{SD(\hat p)}\leq  \frac{0.20-0.30}{0.092} )\\=P(Z\leq -1.087)\\=1-P(Z

**Use the standard normal table for probability.

Thus, the probability that the number of successes is at most 5 is 0.1379.

The exact probability that the number of successes is at most 5 is:

P(S\leq 5)={25\choose 5}(0.30)^{5}91-0.30)^{25-5}=0.1935

The exact probability is more than the approximated probability.

(b)

Compute the probability that the number of successes is at most 7 as follows:

The probability of 5 successes is: p=\frac{7}{25} =0.28

P(\hat p\leq 0.28)=P(\frac{\hat p-E(\hat p)}{SD(\hat p)}\leq  \frac{0.28-0.30}{0.092} )\\=P(Z\leq -0.2174)\\=1-P(Z

**Use the standard normal table for probability.

Thus, the probability that the number of successes is at most 7 is 0.4129.

The exact probability that the number of successes is at most 7 is:

P(S\leq 57)={25\choose 7}(0.30)^{7}91-0.30)^{25-7}=0.5118

The exact probability is more than the approximated probability.

(c)

Compute the probability that the number of successes is at most 9 as follows:

The probability of 5 successes is: p=\frac{9}{25} =0.36

P(\hat p\leq 0.36)=P(\frac{\hat p-E(\hat p)}{SD(\hat p)}\leq  \frac{0.36-0.30}{0.092} )\\=P(Z\leq 0.6522)\\=0.7422

**Use the standard normal table for probability.

Thus, the probability that the number of successes is at most 9 is 0.7422.

The exact probability that the number of successes is at most 9 is:

P(S\leq 9)={25\choose 9}(0.30)^{9}91-0.30)^{25-9}=0.8106

The exact probability is more than the approximated probability.

(d)

Compute the probability that the number of successes is at most 11 as follows:

The probability of 5 successes is: p=\frac{11}{25} =0.44

P(\hat p\leq 0.44)=P(\frac{\hat p-E(\hat p)}{SD(\hat p)}\leq  \frac{0.44-0.30}{0.092} )\\=P(Z\leq 1.522)\\=0.9357

**Use the standard normal table for probability.

Thus, the probability that the number of successes is at most 11 is 0.9357.

The exact probability that the number of successes is at most 11 is:

P(S\leq 11)={25\choose 11}(0.30)^{11}91-0.30)^{25-11}=0.9558

The exact probability is more than the approximated probability.

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Answer:

A) Silvie is incorrect; the sum of a rational number and an irrational number is always irrational

Step-by-step explanation:

The concept of adding rational and irrational numbers states that the result will always be irrational.

Take for instance the addition of 4.5 and 3.142857

4.5 is a rational number because it can easily be expressed in fractions while 3.142857 is irrational because it can't be expressed as a simple fraction.

The addition of these two numbers gives 7.642857

7.642857 is an irrational number.

Similarly, any sum of two numbers (rational and irrational) will always result in irrational number.

Hence, Silvie is incorrect.

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