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AlexFokin [52]
3 years ago
10

Rachael has a rope 5 m 32 cm long that she cut into two pieces. One piece is 249 cm long. How many centimeters long is the other

piece of rope?
Mathematics
1 answer:
satela [25.4K]3 years ago
4 0

1 m = 100 cm, so 5 m = 500 cm plus the 32 cm, so she has a rope that is 532 cm long.

532 - 249 = 283 cm

So the other piece of rope is 283 cm long.

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Find the vertex and length of the latus rectum for the parabola. y=1/6(x-8)^2+6
Ivan

Step-by-step explanation:

If the parabola has the form

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then its vertex is located at the point (h, k). Therefore, the vertex of the parabola

y = \dfrac{1}{6}(x - 8)^2 + 6

is located at the point (8, 6).

To find the length of the parabola's latus rectum, we need to find its focal length <em>f</em>. Luckily, since our equation is in vertex form, we can easily find from the focus (or focal point) coordinate, which is

\text{focus} = (h, k +\frac{1}{4a})

where \frac{1}{4a} is called the focal length or distance of the focus from the vertex. So from our equation, we can see that the focal length <em>f</em> is

f = \dfrac{1}{4(\frac{1}{6})} = \dfrac{3}{2}

By definition, the length of the latus rectum is four times the focal length so therefore, its value is

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5 0
3 years ago
Psychologist Michael Cunningham conducted a survey of university women to see whether, upon graduation, they would prefer to mar
bagirrra123 [75]

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

4 0
3 years ago
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