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Neko [114]
3 years ago
7

Math homework please help new subject dont know it. (trolls = report)

Mathematics
2 answers:
Natasha2012 [34]3 years ago
6 0
Find area on triangle:
A=(1/2)bh
A=(1/2)(8 units)(6 units)
A=(1/2)(48 units^2)
A=24 units^2
Just do the same for the rest of this page following the equation for the area of a triangle A=(1/2)(base)(height)

Find the error:
A=(1/2)bh
100m^2=(1/2)b(20m)
100m^2=(10m)b
10m=b
He forgot to multiply bh by 1/2

Area of parallelogram:
A=bh
A=(3 units)(3 units)
A= 9 units^2
The red line is the height and the bottom of the parallelogram is the base, just count the squares. Imagine cutting the parallelogram at the red line and moving that triangle piece to the other side, it’ll make a rectangle, that’s why the equation is the same.
Shkiper50 [21]3 years ago
6 0
Find area on triangle:

A=(1/2)bh

A=(1/2)(8 units) (6 units)

A=(1/2)(48 units^2)

A=24 units^2

Just do the same for the rest of this page following the equation for the area of a triangle A=(1/2) (base) (height)

Find the error:

A=(1/2)bh

100m^2=(1/2)b(20m)

100m^2=(10m)b

10m=b

He forgot to multiply bh by 1/2

Area of parallelogram:

A=bh

A=(3 units) (3 units)

A= 9 units^2

The red line is the height and the bottom of the parallelogram is the base, just count the squares. Imagine cutting the parallelogram at the red line and moving that triangle piece to the other side, it'll make a rectangle, that's why the equation is the same.
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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}\\\\\\

A. P(x=6)

P(x=6) = \dbinom{25}{6} p^{6}(1-p)^{19}=177100*0.00024*0.00423=0.183\\\\\\

B. P(x≥10)

P(x\geq10)=1-P(x

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\\\\\\

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\\\\\\

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

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

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