The picture won't load, but I would surely help!
Answer:
Sabemos que:
y tenemos que:
Con esto podemos encontrar el valor de a:
8*a/(a+ 4) = 6
8*a = 6*(a + 4) = 6*a + 24
8a - 6a = 24
2a = 24
a = 24/2 = 12.
Tambien sabemos que:
Y de ahí podemos despejar b:
(16*b)/(b + 8) = 12
16*b = 12*(b + 8) = 12b + 96
16b - 12b = 96
4b = 96
b = 96/4 = 24
Entonces tenemos a = 12 y b = 24, y el MH de a y b es:
MH(12,24) = 2*12*24/(12 + 24) = 24*24/36 = 16
Answer:
A
Step-by-step explanation:
How to find the coordinates is by going to the right and going up of where the point is
Hope this helps :)
Data for the question :
102.05 99.85 112.3 97.15 111.23 105.37 105.64 106.5 102.97 107.82 106.36 111.24 107.28 114.14 106.28 106.96 98.25 111.55 107.75 101.02 101.12 97.7 97.66 100.54 115.77 112.91 111.04 112.15 102.87 101.14 107.13 108.56 109.56 103.57 108.68 104.59 116.74 116.22 100.22 103.97 111.2 109.34 115.78 101.59 107.93 104.23 96.25 103.84 102.47 102.96 99.26 101.42 108.58 107.69 99.88 102.71 111.25 99.4 117.04 106.35 110.44 102.34 107.25 107.63 105.2 109.14 115.54 101.51 108.49 112.32 109.27 97.54 102.46 105.94 109.42 111.05 102.63 106.99 102.03 108.84 118.8 108.64 95.35 105.47 104.45 102.15 111.4 108.27 104.82 108.4 109.05 116.11 103.7 121.2 99.62 102.81 109.56 103.35 113.02 103.79
Answer:
Range = 25.35
Variance = 29.46
Standard deviation = 5.43
The variation in price of Prozac is high
Step-by-step explanation:
The range of the data :
Maximum - Minimum.
121.2 - 95.35 = 25.35
The variance, s :
s² = Σ(X - m²) / n - 1
Mean, m = Σx / n
X = individual data point
m = mean of data
n = sample size
Using a calculator of save time and ensure accuracy :
s² = 29.45522
The standard deviation, s
s = sqrt(variance)
s = sqrt(s²)
s = sqrt(29.45522)
s = 5.42726.
The range, variance and standard deviation, all measure the degree of variation in a dataset. The values of these statistical measure obtain for the price of 1 product across different pharmaceutical stores, suggests thatvthe variation in price is high;
With a range of about 25.35 and standard deviation of 5.43
