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3 years ago

PLEAAAAASE help me if you have heart!!

Mathematics

2 answers:

Stolb23 [73]3 years ago

8 0

Answer:

a. 2

b. 3

c. 5

d. 41

e. 26.5

Note:

These answers are correct. Pls, let me know if you have any difficulty answering the questions :)

Have a nice day <3

Kay [80]3 years ago

5 0

2
3
5
40
25

i'm pretty sure this is right, but if not please reply to me and tell me what went wrong thanks

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Of 150 adults selected randomly from one town, 30 of them smoke. Construct a 99% confidence interval for the true percentage of

anyanavicka [17]

Answer:

The 99% confidence interval for the true percentage of all adults in the town that smoke is (0.1159, 0.2841).

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of $\pi$ , and a confidence interval $1-\alpha$ , we have the following confidence interval of proportions.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

z is the zscore that has a pvalue of $1 - \frac{\alpha}{2}$ .

For this problem, we have that:

Of 150 adults selected randomly from one town, 30 of them smoke. This means that $n = 150, p = \frac{30}{150} = 0.2$

99% confidence interval

So $\alpha = 0.01$ , z is the value of Z that has a pvalue of $1 - \frac{0.01}{2} = 0.995$ , so $Z = 2.575$ .

The lower limit of this interval is:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.20 - 2.575\sqrt{\frac{0.20*0.80}{150}} = 0.1159$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.20 + 2.575\sqrt{\frac{0.20*0.80}{150}}{119}} = 0.2841$

The 99% confidence interval for the true percentage of all adults in the town that smoke is (0.1159, 0.2841).

3 0

4 years ago

Suppose you can afford only $200 a month in car payments and your best loan option is a 60-month loan at 3%. How much money coul

g100num [7]

Answer:

$11,130.47

Step-by-step explanation:

The amortization formula can be used. It tells you the monthly payment amount A for some principal P, interest rate r, and n payments.

A = P(r/12)/(1 -(1 +r/12)^(-n))

Filling in your values, we get ...

200 = P(.03/12)/(1 -(1 +.03/12)^-60) = P(.0025)/(1 -1.0025^-60)

P = 200(1 -1.0025^-60)/.0025 ≈ 200×55.6523577

P ≈ 11,130.47

The present value of the loan is $11,130.47.

8 0

3 years ago

Mr. Tolin purchased a used car for $2,816, which was 12% off the asking price. What was the asking price of the used car?

Nina [5.8K]

Answer:

Original price= $3,200

Step-by-step explanation:

Giving the following information:

Purchase price= $2,816

Discount rate= 12% = 0.12

To calculate the original price, we need to use the following formula:

Original price= purchase price / (1 - discount rate)

Original price= 2,816 / 0.88

Original price= $3,200

8 0

3 years ago

We have n = 100 many random variables Xi ’s, where the Xi ’s are independent and identically distributed Bernoulli random variab

777dan777 [17]

Answer:

(a) The distribution of $X=\sum\limits^{n}_{i=1}{X_{i}}$ is a Binomial distribution.

(b) The sampling distribution of the sample mean will be approximately normal.

Step-by-step explanation:

It is provided that random variables $X_{i}$ are independent and identically distributed Bernoulli random variables with p = 0.50.

The random sample selected is of size, n = 100.

(a)

Theorem:

Let $X_{1},\ X_{2},\ X_{3},...\ X_{n}$ be independent Bernoulli random variables, each with parameter p, then the sum of of thee random variables, $X=X_{1}+X_{2}+X_{3}...+X_{n}$ is a Binomial random variable with parameter n and p.

Thus, the distribution of $X=\sum\limits^{n}_{i=1}{X_{i}}$ is a Binomial distribution.

(b)

According to the Central Limit Theorem if we have an unknown population with mean μ and standard deviation σ and appropriately huge random samples (n > 30) are selected from the population with replacement, then the distribution of the sample mean will be approximately normally distributed.

The sample size is large, i.e. n = 100 > 30.

So, the sampling distribution of the sample mean will be approximately normal.

The mean of the distribution of sample mean is given by,

$\mu_{\bar x}=\mu=p=0.50$