Given equation : n(17+x)=34x−r.
We need to solve it for x.
Distributing n over (17+x) on left side, we get
17n + nx = 34x - r.
Adding r on both sides, we get
17n+r + nx = 34x - r+r.
17n + r + nx = 34x.
Subtracting nx from both sides, we get
17n + r + nx-nx = 34x-nx
17n + r = 34x -nx.
Factoring out gcf x on right side, we get
17x + r = x(34-n).
Dividing both sides by (34-n), we get
Answer:
The volume of cone v= 113.04 cm³.
Step-by-step explanation:
Formula used to find volume of cone is:
where r is the radius and h is the height of cone.
In the given question we are given:
height h= 6 cm
diameter d= 6 cm
We know radius r = d/2 = 6/2 = 3 cm
so, volume will be:
So, the volume of cone v= 113.04 cm³ given height h= 6cm and radius r= 3cm
Answer:
a = -1 , b = 4 , c = 3 , d = 1
Step-by-step explanation:
The general rules of exponent:
===================================================
Given:
The right hand side = (-xy)³ (xz) = (-1)³ x³ y³ x z = (-1) x⁴ y³ z
Comparing the right hand side with the left hand side
∴ a = -1
∴ b = 4
∴ c = 3
∴ d = 1
Answer:
Step-by-step explanation:
Hello!
Given the variable
X: breakdown voltage of an insulating liquid under certain conditions.
a)
Graphic in attachment.
Box: The median is closer to the third quartile than the first quartile showing skewness to the left.
The black dot in the middle of the box represents the sample mean (X[bar]) as you can see it is also moved to the right side of the distribution, affected by the presence of an outlier.
Whiskers:
The upper whisker is smaller than the lower one and there is a value that can be considered an outlier.
Altogether you can say that the distribution of the data set is skewed to the left.
b)
To estimate the population mean you need the population to be at least normal. The box plot shows that the distribution is not symmetrical but, considering that the sample size is n= 48 you can apply the Central Limit Theorem and approximate the sampling distribution to normal.
X[bar]≈N(μ;σ²)
As the distribution is approximate, using the sample standard deviation (since we don't know the population value) for the distribution standard deviation is also valid ⇒
X[bar] ± *
∑X= 2626 ∑X²= 144950 n= 48
X[bar]= ∑X/n= 2626/48= 54.71
S²= 1/n[∑X²- (∑X)²/n]= 1/47[144950-(2626)²/48]= 27.36
S= 5.23
54.71 ± 1.96 *
[53.22; 56.18]kV
c)
95% CI
amplitude a=2 kV i.e. semi amplitude d= 1 kV
The semi amplitude or margin of error for the CI can be calculated as:
d= *
From this formula you can clear the sample size
