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Lynna [10]
2 years ago
10

Carl Cornfield has been wondering whether he should plant white corn this year. He decides to sample 100 customers. 38% say they

would purchase white corn. Carl wants to have a 95% confidence interval for this proportion. The confidence interval is from 33% to 43% (to the nearest percent). Truth Or False
Mathematics
2 answers:
krek1111 [17]2 years ago
8 0

The answer is false

This should be the interval:

.285 to .475

Hope this helps

-AaronWiseIsBae

kirza4 [7]2 years ago
6 0

Answer is false

Here sample size = 100

Sample propotion = 38%

i.e. p =0.38 and q = 1-p =0.62

p follows a normal with mean = 0.38 and std error = \sqrt{}\frac{pq}{n}  =0.0485

For 95% z value = 1.96

Hence margin of error = 1.96 (std error) = 0.0951

Confidence interval = (0.38-0.0951, 0.38+0.0951)

=(0.2949, 0.4751)

= (29.49%, 47.51%)

Hence given confidence interval is from 33% to 43% (to the nearest percent).

False

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Answer:

After 6 months, the interest generated by the investment would be $21.10.

Step-by-step explanation:

To determine the interest that Preston McCord could earn by investing $ 1,400 in an account that pays 3.2% annual interest compounded monthly, leaving said money invested for a period of 6 months, it is necessary to perform the following calculation:

X = 1,400 x (1 + 0.032 / 6) ^ 0.5x6

X = 1,421.10

1421.10 - 1400 = 21.10

Thus, after 6 months, the interest generated by the account will be $ 21.10.

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2 years ago
USE THE IMAGE ATTACHED BELOW please help me with my work answer it correctly I HAVE SO MUCH WORK DURING QUARANTINE
andrezito [222]

Answer:

Question 1:

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Question 2:

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8 0
3 years ago
Read 2 more answers
Of the entering class at a​ college, ​% attended public high​ school, ​% attended private high​ school, and ​% were home schoole
Veronika [31]

Answer:

(a) The probability that the student made the​ Dean's list is 0.1655.

(b) The probability that the student came from a private high school, given that the student made the Dean's list is 0.2411.

(c) The probability that the student was not home schooled, given that the student did not make the Dean's list is 0.9185.

Step-by-step explanation:

The complete question is:

Of the entering class at a college, 71% attended public high school, 21% attended private high school, and 8% were home schooled. Of those who attended public high school, 16% made the Dean's list, 19% of those who attended private high school made the Dean's list, and 15% of those who were home schooled made the Dean's list.

a) Find the probability that the student made the Dean's list.

b) Find the probability that the student came from a private high school, given that the student made the Dean's list.

c) Find the probability that the student was not home schooled, given that the student did not make the Dean's list.

Solution:

Denote the events as follows:

<em>A</em> = a student attended public high school

<em>B</em> = a student attended private high school

<em>C</em> = a student was home schooled

<em>D</em> = a student made the Dean's list

The provided information is as follows:

P (A) = 0.71

P (B) = 0.21

P (C) = 0.08

P (D|A) = 0.16

P (D|B) = 0.19

P (D|C) = 0.15

(a)

The law of total probability states that:

P(X)=\sum\limits_{i} P(X|Y_{i})\cdot P(Y_{i})

Compute the probability that the student made the​ Dean's list as follows:

P(D)=P(D|A)P(A)+P(D|B)P(B)+P(D|C)P(C)

         =(0.16\times 0.71)+(0.19\times 0.21)+(0.15\times 0.08)\\=0.1136+0.0399+0.012\\=0.1655

Thus, the probability that the student made the​ Dean's list is 0.1655.

(b)

Compute the probability that the student came from a private high school, given that the student made the Dean's list as follows:

P(B|D)=\frac{P(D|B)P(B)}{P(D)}

             =\frac{0.21\times 0.19}{0.1655}\\\\=0.2410876\\\\\approx 0.2411

Thus, the probability that the student came from a private high school, given that the student made the Dean's list is 0.2411.

(c)

Compute the probability that the student was not home schooled, given that the student did not make the Dean's list as follows:

P(C^{c}|D^{c})=1-P(C|D^{c})

               =1-\frac{P(D^{c}|C)P(C)}{P(D^{c})}\\\\=1-\frac{(1-P(D|C))\times P(C)}{1-P(D)}\\\\=1-\frac{(1-0.15)\times 0.08}{(1-0.1655)}\\\\=1-0.0815\\\\=0.9185

Thus, the probability that the student was not home schooled, given that the student did not make the Dean's list is 0.9185.

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butalik [34]

t³-t²-2t/t -5 )/(t²-t-2)

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7 0
3 years ago
Help please!!!!!!!!!!!!!!?!?
Ahat [919]

0.25 = 1/4, 0.8 : 4 = 0.2

4 0
3 years ago
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