First, subtract '11' from each side.
Second, subtract '29 - 11' to get '18'.
Third, since we are solving for f, we have to get it by itself. Multiply each side by 'f'.
Fourth, once again, we need to get 'f' alone. Divide both sides by 18.
Fifth, now that we have 'f' by itself, we can simplify the current fraction. To do so, we need to start off with listing the factors of 4 and 18 and find the greatest common factor (GCF).
Factors of 4: 1, 2, 4
Factors of 18: 1, 2, 3, 6, 9, 18
Out of the listed factors, 1 and 2 are the common factors, and since 2 is the highest number out of them, it is considered the greatest common factor. The GCF is 2.
Sixth, divide the numerator and denominator by the GCF (2).
Seventh, we can now rewrite our fraction in simplest form and switch sides.
Answer in fraction form:
Answer in decimal form:
what is the maximum, minimum, quartile 1, median, quartile 3, range, interquartlie range of these numbers " 46,48,50,52, and 54"
Gekata [30.6K]
Min=46
Max=54
1 quartile= 48
Median=50
3 quartile=52
46/48 percent is 95.83%
Answer:
17.9 m/s
Step-by-step explanation:
Volume of the slick = 0.5 x π r² h--------------------------------- (1)
Where r = radius of slick
h = thickness of slick, 10⁻⁶m
If 0.5m³ of oil leaked, then the radius of the semicircular slick can be calculated from equation (1)
V = 0.5 x π r² h
0.5= 0.5 x π x r² x 10⁻⁶
r² = 10⁶/ π
r = 10³/√π
dV/dt = πrh dr/dt + 0.5π r² dh/dt----------------------------------- (2)
Asumming the film thickness is constant , equation (2) becomes
dV/dt = πrh dr/dt-------------------------------- (3)
dV/dt = 0.1m³/day
r= 10³/√π
dr/dt= rate of expansion of the slick
Substituting into (3);
0.1 = π x 10³/√π x 10⁻⁶ x dr/dt
dr/dt = 0.1 x 10⁶/ ( π x 10³/√π)
= 17.9479 m/s
≅ 17.9 m/s
Answer:
Standard error of mean = 689
Step-by-step explanation:
We are given the following information in the question:
Mean, μ = $28,520
Standard Deviation, σ = $5600
Mean of sampling distribution =
As per Central Limit Theorem, if the sample size is large enough, then the sampling distribution of the sample means follow approximately a normal distribution.
Sample size, n = 66
Since the sample size is large, we can use normal distribution for approximation.
Standard error of mean =
Answer:
3:6
3x2=6
Scale factor = 2
Step-by-step explanation:
