Find the volume 3yd 6yd 7yd

Find the exponential generating function for the number of alphanumeric strings of length n n formed from the 26 26 uppercase le

lana66690 [7]

Answer:

the answer is contained in the attachment

Step-by-step explanation:

for complete explanation kindly check the attachment.

5 0

3 years ago

The National Center for Education Statistics surveyed a random sample of 4400 college graduates about the lengths of time requir

Paha777 [63]

Answer:

95% confidence interval for the mean time required to earn a bachelor’s degree by all college students is [5.10 years , 5.20 years].

Step-by-step explanation:

We are given that the National Center for Education Statistics surveyed a random sample of 4400 college graduates about the lengths of time required to earn their bachelor’s degrees. The mean was 5.15 years and the standard deviation was 1.68 years respectively.

Firstly, the pivotal quantity for 95% confidence interval for the population mean is given by;

P.Q. = $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, $\bar X$ = sample mean time = 5.15 years

$\sigma$ = sample standard deviation = 1.68 years

n = sample of college graduates = 4400

$\mu$ = population mean time

Here for constructing 95% confidence interval we have used One-sample z test statistics although we are given sample standard deviation because the sample size is very large so at large sample values t distribution also follows normal.

So, 95% confidence interval for the population mean, $\mu$ is ;

P(-1.96 < N(0,1) < 1.96) = 0.95 {As the critical value of z at 2.5%

level of significance are -1.96 & 1.96}

P(-1.96 < $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ < 1.96) = 0.95

P( $-1.96 \times {\frac{\sigma}{\sqrt{n} } }$ < ${\bar X-\mu}$ < $1.96 \times {\frac{\sigma}{\sqrt{n} } }$ ) = 0.95

P( $\bar X-1.96 \times {\frac{\sigma}{\sqrt{n} } }$ < $\mu$ < $\bar X+1.96 \times {\frac{\sigma}{\sqrt{n} } }$ ) = 0.95

95% confidence interval for $\mu$ = [ $\bar X-1.96 \times {\frac{\sigma}{\sqrt{n} } }$ , $\bar X+1.96 \times {\frac{\sigma}{\sqrt{n} } }$ ]

= [ $5.15-1.96 \times {\frac{1.68}{\sqrt{4400} } }$ , $5.15+1.96 \times {\frac{1.68}{\sqrt{4400} } }$ ]

= [5.10 , 5.20]

Therefore, 95% confidence interval for the mean time required to earn a bachelor’s degree by all college students is [5.10 years , 5.20 years].

8 0

3 years ago

Craig is designing a new cylinder-shaped can for a pet food company. He wants the height of the can to be 40 mm and the radius o

devlian [24]

The height of the can for this case is given by:
h = 40 mm
The radius of the can is given by:
r = (1 3/8) * h
Substituting values:
r = (1 3/8) * 40
r = 55 mm
Therefore, the diameter of the can is:
d = 2 * r
d = 2 * 55
d = 110 mm
Answer:
The diameter of the can will be:
d = 110 mm

3 0

3 years ago

According to a study done by Nick Wilson of Otago University Wellington, the probability a randomly selected individual will not

Lorico [155]

Answer and Step-by-step explanation:

From the question statement we get know that it is Binomial distribution because there are only two possible outcomes so we need to use Binomial Probability Distribution for this question.

Formula for the Binomial Probability Distribution:

P(X)= p^x q^(n-x)

Where,

C_x^n=n!/(n-x)!x! (i.e. combination)
x= total number of successes
p=probability of success (p=1-q)
q=probability of failure (q=1-p)
n=number of trials
P(X)= probability of total number of successes

Answer and explanation for each part of the question are as follow:

a.What is the probability that among 10 randomly observed individuals exactly 4 do not cover their mouth when sneezing?

Solution:

Given that

n=10

p=0.267 (because p is the probability of success which is “number of individuals not covering their mouths when sneezing” in the question)

q=1-0.267=0.733

x=4 (number of successes i.e. individuals not covering their mouths)

C_x^n=n!/(n-x)!x!=10!/(10-4)!4!=210

P(X)=C_x^n p^x q^(n-x)=210×〖(0.267)〗^4×〖0.733〗^(10-4)

P(X)=210×0.00508×0.155

P(X)=0.165465

b. What is the probability that among 10 randomly observed individuals fewer than 3 do not cover their mouth when sneezing?

Solution:

Given that

n=10

p=0.267 (because p is the probability of success which is “number of individuals not covering their mouths when sneezing” in the question)

q=1-0.267=0.733

x=3 (number of successes i.e. individuals not covering their mouths)

C_x^n=n!/(n-x)!x!=10!/(10-3)!3!=120

P(X)=C_x^n p^x q^(n-x)=120×(0.267)^3×〖0.733〗^(10-3)

P(X)=120×0.01903×0.1136

P(X)=0.25962

c. Would you be surprised if, after observing 18 individuals, fewer than half covered their mouth when sneezing? why?

Solution:

Given that

n=18

p=0.267 (because p is the probability of success which is “number of individuals not covering their mouths when sneezing” in the question)

q=1-0.267=0.733

x=9 (x is the number of successes “number of individuals not covering their mouths when sneezing”, if less than half cover their mouth then more than half will not cover), so let x=9

C_x^n=n!/(n-x)!x!=18!/(18-9)!9!=48620

P(X)=48620×(0.267)^9×〖0.733〗^(18-9)

P(X)=48620×0.00000689×0.0610

P(X)=0.020

Yes, I am surprised that probability of less than 9 individuals covering their mouth when sneezing is 0.020. Which is extremely is small.

3 0

3 years ago

A new car is purchased for 19900 dollars. The value of the car depreciates at 7.5% per year. What will be the value of the car a

bogdanovich [222]

Answer:

$10665,64

Step-by-step explanation:

S=19900*(1-(7,5/100))^8=10665, 64

6 0

3 years ago

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