Answer:
we say for μ = 50.00 mm we be 95% confident that machine calibrated properly with ( 49.926757 , 50.033243 )
Step-by-step explanation:
Given data
n=29
mean of x = 49.98 mm
S = 0.14 mm
μ = 50.00 mm
Cl = 95%
to find out
Can we be 95% confident that machine calibrated properly
solution
we know from t table
t at 95% and n -1 = 29-1 = 28 is 2.048
so now
Now for 95% CI for mean is
(x - 2.048 × S/√n , x + 2.048 × S/√n )
(49.98 - 2.048 × 0.14/√29 , 49.98 + 2.048 × 0.14/√29 )
( 49.926757 , 50.033243 )
hence we say for μ = 50.00 mm we be 95% confident that machine calibrated properly with ( 49.926757 , 50.033243 )
Answer: The radius of the circular rug is 6 meters.
Step-by-step explanation:
Hi, to answer this question we have to apply the next formula:
<em>Area of a circle (A) = pi r²
</em>
Where r is the radius:
Replacing with the values given:
36 pi=pi r²
Dividing both sides by pi:
36 = r²
√36= r
r = 6 meters
So, the radius of the circular rug is 6 meters.
Feel free to ask for more if needed or if you did not understand something.
Answer:someone answer the question already
Step-by-step explanation:
(9 × 10) + (20x + 21) = 571
90 + 20x + 21 = 571
111 + 20x = 571
- 111
20x = 460
÷ 20
x = 23
I hope this helps!
Answer:
The mean and the standard deviation of the number of students with laptops are 1.11 and 0.836 respectively.
Step-by-step explanation:
Let <em>X</em> = number of students who have laptops.
The probability of a student having a laptop is, P (X) = <em>p</em> = 0.37.
A random sample of <em>n</em> = 30 students is selected.
The event of a student having a laptop is independent of the other students.
The random variable <em>X</em> follows a Binomial distribution with parameters <em>n</em> and <em>p</em>.
The mean and standard deviation of a binomial random variable <em>X</em> are:
Compute the mean of the random variable <em>X</em> as follows:
The mean of the random variable <em>X</em> is 1.11.
Compute the standard deviation of the random variable <em>X</em> as follows:
The standard deviation of the random variable <em>X</em> is 0.836.
