Answer:
A.The probability that exactly six of Nate's dates are women who prefer surgeons is 0.183.
B. The probability that at least 10 of Nate's dates are women who prefer surgeons is 0.0713.
C. The expected value of X is 6.75, and the standard deviation of X is 2.17.
Step-by-step explanation:
The appropiate distribution to us in this model is the binomial distribution, as there is a sample size of n=25 "trials" with probability p=0.25 of success.
With these parameters, the probability that exactly k dates are women who prefer surgeons can be calculated as:
![P(x=k) = \dbinom{n}{k} p^{k}(1-p)^{n-k}\\\\\\P(x=k) = \dbinom{25}{k} 0.25^{k} 0.75^{25-k}\\\\\\](https://tex.z-dn.net/?f=P%28x%3Dk%29%20%3D%20%5Cdbinom%7Bn%7D%7Bk%7D%20p%5E%7Bk%7D%281-p%29%5E%7Bn-k%7D%5C%5C%5C%5C%5C%5CP%28x%3Dk%29%20%3D%20%5Cdbinom%7B25%7D%7Bk%7D%200.25%5E%7Bk%7D%200.75%5E%7B25-k%7D%5C%5C%5C%5C%5C%5C)
A. P(x=6)
![P(x=6) = \dbinom{25}{6} p^{6}(1-p)^{19}=177100*0.00024*0.00423=0.183\\\\\\](https://tex.z-dn.net/?f=P%28x%3D6%29%20%3D%20%5Cdbinom%7B25%7D%7B6%7D%20p%5E%7B6%7D%281-p%29%5E%7B19%7D%3D177100%2A0.00024%2A0.00423%3D0.183%5C%5C%5C%5C%5C%5C)
B. P(x≥10)
![P(x\geq10)=1-P(x](https://tex.z-dn.net/?f=P%28x%5Cgeq10%29%3D1-P%28x%3C10%29%3D1-%5Csum_%7Bi%3D0%7D%5E9P%28x%3Di%29%5C%5C%5C%5C%5C%5C)
![P(x=0) = \dbinom{25}{0} p^{0}(1-p)^{25}=1*1*0.0008=0.0008\\\\\\P(x=1) = \dbinom{25}{1} p^{1}(1-p)^{24}=25*0.25*0.001=0.0063\\\\\\P(x=2) = \dbinom{25}{2} p^{2}(1-p)^{23}=300*0.0625*0.0013=0.0251\\\\\\P(x=3) = \dbinom{25}{3} p^{3}(1-p)^{22}=2300*0.0156*0.0018=0.0641\\\\\\P(x=4) = \dbinom{25}{4} p^{4}(1-p)^{21}=12650*0.0039*0.0024=0.1175\\\\\\P(x=5) = \dbinom{25}{5} p^{5}(1-p)^{20}=53130*0.001*0.0032=0.1645\\\\\\P(x=6) = \dbinom{25}{6} p^{6}(1-p)^{19}=177100*0.0002*0.0042=0.1828\\\\\\](https://tex.z-dn.net/?f=P%28x%3D0%29%20%3D%20%5Cdbinom%7B25%7D%7B0%7D%20p%5E%7B0%7D%281-p%29%5E%7B25%7D%3D1%2A1%2A0.0008%3D0.0008%5C%5C%5C%5C%5C%5CP%28x%3D1%29%20%3D%20%5Cdbinom%7B25%7D%7B1%7D%20p%5E%7B1%7D%281-p%29%5E%7B24%7D%3D25%2A0.25%2A0.001%3D0.0063%5C%5C%5C%5C%5C%5CP%28x%3D2%29%20%3D%20%5Cdbinom%7B25%7D%7B2%7D%20p%5E%7B2%7D%281-p%29%5E%7B23%7D%3D300%2A0.0625%2A0.0013%3D0.0251%5C%5C%5C%5C%5C%5CP%28x%3D3%29%20%3D%20%5Cdbinom%7B25%7D%7B3%7D%20p%5E%7B3%7D%281-p%29%5E%7B22%7D%3D2300%2A0.0156%2A0.0018%3D0.0641%5C%5C%5C%5C%5C%5CP%28x%3D4%29%20%3D%20%5Cdbinom%7B25%7D%7B4%7D%20p%5E%7B4%7D%281-p%29%5E%7B21%7D%3D12650%2A0.0039%2A0.0024%3D0.1175%5C%5C%5C%5C%5C%5CP%28x%3D5%29%20%3D%20%5Cdbinom%7B25%7D%7B5%7D%20p%5E%7B5%7D%281-p%29%5E%7B20%7D%3D53130%2A0.001%2A0.0032%3D0.1645%5C%5C%5C%5C%5C%5CP%28x%3D6%29%20%3D%20%5Cdbinom%7B25%7D%7B6%7D%20p%5E%7B6%7D%281-p%29%5E%7B19%7D%3D177100%2A0.0002%2A0.0042%3D0.1828%5C%5C%5C%5C%5C%5C)
![P(x=7) = \dbinom{25}{7} p^{7}(1-p)^{18}=480700*0.000061*0.005638=0.1654\\\\\\P(x=8) = \dbinom{25}{8} p^{8}(1-p)^{17}=1081575*0.000015*0.007517=0.1241\\\\\\P(x=9) = \dbinom{25}{9} p^{9}(1-p)^{16}=2042975*0.000004*0.010023=0.0781\\\\\\](https://tex.z-dn.net/?f=P%28x%3D7%29%20%3D%20%5Cdbinom%7B25%7D%7B7%7D%20p%5E%7B7%7D%281-p%29%5E%7B18%7D%3D480700%2A0.000061%2A0.005638%3D0.1654%5C%5C%5C%5C%5C%5CP%28x%3D8%29%20%3D%20%5Cdbinom%7B25%7D%7B8%7D%20p%5E%7B8%7D%281-p%29%5E%7B17%7D%3D1081575%2A0.000015%2A0.007517%3D0.1241%5C%5C%5C%5C%5C%5CP%28x%3D9%29%20%3D%20%5Cdbinom%7B25%7D%7B9%7D%20p%5E%7B9%7D%281-p%29%5E%7B16%7D%3D2042975%2A0.000004%2A0.010023%3D0.0781%5C%5C%5C%5C%5C%5C)
![P(x\geq10)=1-(0.0008+0.0063+0.0251+0.0641+0.1175+0.1645+0.1828+0.1654+0.1241+0.0781)\\\\P(x\geq10)=1-0.9287=0.0713](https://tex.z-dn.net/?f=P%28x%5Cgeq10%29%3D1-%280.0008%2B0.0063%2B0.0251%2B0.0641%2B0.1175%2B0.1645%2B0.1828%2B0.1654%2B0.1241%2B0.0781%29%5C%5C%5C%5CP%28x%5Cgeq10%29%3D1-0.9287%3D0.0713)
C. The expected value (mean) and standard deviation of this binomial distribution can be calculated as:
![E(x)=\mu=n\cdot p=25\cdot 0.25=6.25\\\\\sigma=\sqrt{np(1-p)}=\sqrt{25\cdot 0.25\cdot 0.75}=\sqrt{4.69}\approx2.17](https://tex.z-dn.net/?f=E%28x%29%3D%5Cmu%3Dn%5Ccdot%20p%3D25%5Ccdot%200.25%3D6.25%5C%5C%5C%5C%5Csigma%3D%5Csqrt%7Bnp%281-p%29%7D%3D%5Csqrt%7B25%5Ccdot%200.25%5Ccdot%200.75%7D%3D%5Csqrt%7B4.69%7D%5Capprox2.17)
Answer:
B
Step-by-step explanation:
To evaluate the function, substitute x = -3. Since it is a negative exponent, it will create the reciprocal of the base.
![y = 3 (\frac{1}{2})^x\\\\y = 3 (\frac{1}{2})^{-3}\\\\y = 3 (\frac{2}{1})^3\\y = 3 (2^3)\\y = 3(8)\\y=24](https://tex.z-dn.net/?f=y%20%3D%203%20%28%5Cfrac%7B1%7D%7B2%7D%29%5Ex%5C%5C%5C%5Cy%20%3D%203%20%28%5Cfrac%7B1%7D%7B2%7D%29%5E%7B-3%7D%5C%5C%5C%5Cy%20%3D%203%20%28%5Cfrac%7B2%7D%7B1%7D%29%5E3%5C%5Cy%20%3D%203%20%282%5E3%29%5C%5Cy%20%3D%203%288%29%5C%5Cy%3D24)
(1/3)/(5)= 1/3 * 1/5 = 1/15 of a chocolate bar each friend gets
Answer:
C. 0.1, 0.2, 0.7, 0.9 this is the answer
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