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Vesnalui [34]
1 year ago
11

g Students conducted a survey and found out that 36% of their peers on campus had tattoos but only 4% of their peers were smoker

s. If 100 students were surveyed, can these students use the Normal approximation to construct a confidence interval for the proportion of students in the population who are smokers? No, because either n p np or n ( 1 − p ) n(1−p) are greater than 15. Yes, because both n p np and n ( 1 − p ) n(1−p) are greater than 15. Yes, because both n p np and n ( 1 − p ) n(1−p) are less than 15. No, because either n p np or n ( 1 − p ) n(1−p) are less than 15.
Mathematics
1 answer:
IgorC [24]1 year ago
5 0

Answer:

D) No, because either np or n(1−p) are less than 15.

Step-by-step explanation:

Percentage of students who had tattoos = 36%

Percentage of students who were smokers = 4%

Sample size = n = 100

The condition to use the Normal distribution as an approximation to construct the confidence interval for population proportion is:

Both np and n(1-p) must be equal to or greater than 15.

Since, we are interested in smokers only, so p = 4% = 0.04

np = 100  x 0.04 = 4

n (1 - p) = 100 x 0.96 = 96

Since, np < 15, we cannot use the Normal distribution as an approximation here.

Therefore, the correct answer is:

No, because either np or n(1−p) are less than 15.

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A dartboard has 10 equally sized slices numbered from 1 to 10. Some are grey and some are white. The slices numbered 1,2 ,3 ,5 ,
hram777 [196]

Answer:

P(X) is 7/10

P(not X) is 3/10

Step-by-step explanation:

The total is 10 so thats ur denominator

The probability of landing on a grey area is 7/10

The probability of not landing on a grey area is 3/10

Hope this helps

5 0
2 years ago
Work out the percentage change to 2 decimal places when a price of £87.95 is decreased to £70.
kap26 [50]

Answer:

20.41%

Step-by-step explanation:

87.95 - 70 = 17.95

17.95 / 87.95= 0.20409323479

0.20409323479 x 100 = 20.409323479

round it

= 20.41%

5 0
2 years ago
Read 2 more answers
Let P and Q be polynomials with positive coefficients. Consider the limit below. lim x→[infinity] P(x) Q(x) (a) Find the limit i
jenyasd209 [6]

Answer:

If the limit that you want to find is \lim_{x\to \infty}\dfrac{P(x)}{Q(x)} then you can use the following proof.

Step-by-step explanation:

Let P(x)=a_{n}x^{n}+a_{n-1}x^{n-1}+\cdots+a_{1}x+a_{0} and Q(x)=b_{m}x^{m}+b_{m-1}x^{n-1}+\cdots+b_{1}x+b_{0} be the given polinomials. Then

\dfrac{P(x)}{Q(x)}=\dfrac{x^{n}(a_{n}+a_{n-1}x^{-1}+a_{n-2}x^{-2}+\cdots +a_{2}x^{-(n-2)}+a_{1}x^{-(n-1)}+a_{0}x^{-n})}{x^{m}(b_{m}+b_{m-1}x^{-1}+b_{n-2}x^{-2}+\cdots+b_{2}x^{-(m-2)}+b_{1}x^{-(m-1)}+b_{0}x^{-m})}=x^{n-m}\dfrac{a_{n}+a_{n-1}x^{-1}+a_{n-2}x^{-2}+\cdots +a_{2}x^{-(n-2)}+a_{1}x^{-(n-1)})+a_{0}x^{-n}}{b_{m}+b_{m-1}x^{-1}+b_{n-2}x^{-2}+\cdots+b_{2}x^{-(m-2)}+b_{1}x^{-(m-1)}+b_{0}x^{-m}}

Observe that

\lim_{x\to \infty}\dfrac{a_{n}+a_{n-1}x^{-1}+a_{n-2}x^{-2}+\cdots +a_{2}x^{-(n-2)}+a_{1}x^{-(n-1)})+a_{0}x^{-n}}{b_{m}+b_{m-1}x^{-1}+b_{n-2}x^{-2}+\cdots+b_{2}x^{-(m-2)}+b_{1}x^{-(m-1)}+b_{0}x^{-m}}=\dfrac{a_{n}}{b_{m}}

and

\lim_{x\to \infty} x^{n-m}=\begin{cases}0& \text{if}\,\, nm\end{cases}

Then

\lim_{x\to \infty}=\lim_{x\to \infty}x^{n-m}\dfrac{a_{n}+a_{n-1}x^{-1}+a_{n-2}x^{-2}+\cdots +a_{2}x^{-(n-2)}+a_{1}x^{-(n-1)}+a_{0}x^{-n}}{b_{m}+b_{m-1}x^{-1}+b_{n-2}x^{-2}+\cdots+b_{2}x^{-(m-2)}+b_{1}x^{-(m-1)}+b_{0}x^{-m}}=\begin{cases}0 & \text{if}\,\, nm \end{cases}

3 0
1 year ago
In the year 2000, the population of a city was 600,000 citizens. The population increases at a rate of 1.8% per year.
Alona [7]
Y=600000(1+1.8%)^(x-2000)
So the population in 2012
Y=600,000×(1+0.018)^(2,012−2,000)
Y=743,232
6 0
1 year ago
Read 2 more answers
An urn contains 12 balls, five of which are red. Selection of a red ball is desired and is therefore considered to be a success.
SVEN [57.7K]

Answer:

2/10

Step-by-step explanation:

8 0
2 years ago
Read 2 more answers
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