Set A's standard deviation exceeds that of Set B. To explain, standard deviation reflects variation within data sets. Generally, a dataset with a narrower range will exhibit a smaller standard deviation. For Set A, the range is 25-1 = 24, while for Set B, it's 18-8 = 10. Given that Set A's range is bigger, we would anticipate its standard deviation to also be larger. Standard deviation is calculated as the square root of the average of the squared deviations from the mean. In Set A, the deviations are ±12, ±11, ±10, whereas Set B's deviations are ±5, ±3, ±1. We can reasonably conclude that the value for Set A will be greater without computing the RMS difference. Thus, Set A's standard deviation is larger compared to Set B.
Respuesta:
la cantidad de elemento restante después de 14 minutos = 7.091 g =~ 10 g
Explicación paso a paso:
Después de cada minuto, la cantidad que queda será
(100 - 26.9) % es decir, 73.1 %
lo que equivale a 0.731 veces la cantidad inicial.
Si el tiempo transcurrido se representa como t, la función f(t) indica la masa del elemento restante, nuestra ecuación será
f(t) = 570(0.731) ^ t
t= 14 minutos
f(14) = 570 (0.731) ^ 14
= 7.091 g =~ 10 g
Response:
![f(x)=4\sqrt[3]{16}^{2x}](https://tex.z-dn.net/?f=f%28x%29%3D4%5Csqrt%5B3%5D%7B16%7D%5E%7B2x%7D)
Detailed explanation:
You're likely in search of a function with a base that can be simplified to...
![4\sqrt[3]{4}\approx 6.3496](https://tex.z-dn.net/?f=4%5Csqrt%5B3%5D%7B4%7D%5Capprox%206.3496)
The functions you seem to be considering appear to be...
![f(x)=2\sqrt[3]{16}^x\approx 2\cdot2.5198^x\\\\f(x)=2\sqrt[3]{64}^x=2\cdot 4^x\\\\f(x)=4\sqrt[3]{16}^{2x}\approx 4\cdot 6.3496^x\ \leftarrow\text{ this one}\\\\f(x)=4\sqrt[3]{64}^{2x}=4\cdot 16^x](https://tex.z-dn.net/?f=f%28x%29%3D2%5Csqrt%5B3%5D%7B16%7D%5Ex%5Capprox%202%5Ccdot2.5198%5Ex%5C%5C%5C%5Cf%28x%29%3D2%5Csqrt%5B3%5D%7B64%7D%5Ex%3D2%5Ccdot%204%5Ex%5C%5C%5C%5Cf%28x%29%3D4%5Csqrt%5B3%5D%7B16%7D%5E%7B2x%7D%5Capprox%204%5Ccdot%206.3496%5Ex%5C%20%5Cleftarrow%5Ctext%7B%20this%20one%7D%5C%5C%5C%5Cf%28x%29%3D4%5Csqrt%5B3%5D%7B64%7D%5E%7B2x%7D%3D4%5Ccdot%2016%5Ex)
It looks like the third option is the one that fits your requirements.
1) For the expected number in 200 restaurants where exactly 8 customers use the drive-through, it is 20.66. Similarly, for exactly 9 customers, the expected count is 13.76. To compute these expectations, we utilize the Poisson distribution which defines the success probability X, considering a certain operational rate for the drive-through services during specified times.
A matching complex for 2+3i is required. The conjugate is 2-3i, leading to the factors (x-2-3i)(x-2+3i)=(x²-4x+4+9)=x²-4x+13. The resulting polynomial is (x-4)(x+8)(x²-4x+13)=(x²+4x-32)(x²-4x+13)=x⁴-4x³+13x²+4x³-16x²+52x-32x²+128x-416, resulting in the 4th degree polynomial: x⁴-35x²+180x-416.
