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natita [175]
3 years ago
7

Using geometry, calculate the volume of the solid under z=64−x2−y2and over the circular disk x2+y2≤64

Mathematics
1 answer:
Anarel [89]3 years ago
3 0
Perhaps you mean to find the volume under z=\sqrt{64-x^2-y^2}? It seems like you have to rely on calculus otherwise (in the case that you indeed mean z=64-x^2-y^2).

Assuming this, note that z=\sqrt{64-x^2-y^2} describes the top half of a sphere with radius 8, which means the volume below this surface over the disk x^2+y^2\le64 is simply half the volume of the sphere. Thus the volume is

\dfrac43\pi8^3=\dfrac{2048}3\pi
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Answer:

The probability that 8 mice are​ required is 0.2428.

Step-by-step explanation:

Given : A scientist inoculates​ mice, one at a​ time, with a disease germ until he finds 3 that have contracted the disease. If the probability of contracting the disease is two sevenths.

To find : What is the probability that 8 mice are​ required? The probability that that 8 mice are required is nothing ?

Solution :

Applying binomial distribution,

P(X=r)=^nC_r p^rq^{n-r}

Where, p is the probability of success p=\frac{2}{7}

q is the probability of failure q=1-p, q=1-\frac{2}{7}=\frac{5}{7}

n is total number of trials n=8

r=3

Substitute the values,

P(X=3)=^8C_3 (\frac{2}{7})^3 (\frac{5}{7})^{8-3}

P(X=3)=\frac{8!}{3!5!}\times \frac{8}{343}\times (\frac{5}{7})^{5}

P(X=3)=\frac{8\times 7\times 6}{3\times 2\times 1}\times \frac{8}{343}\times \frac{3125}{16807}

P(X=3)=0.2428

Therefore, the probability that 8 mice are​ required is 0.2428.

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2+d=2-3(d-5)-2 Pls show work
defon
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First, cancel 2 on both sides. / Your problem should look like: d = -3(d - 5) - 2
Second, expand. / Your problem should look like: d = -3d + 15 - 2
Third, simplify -3d + 15 - 2 to get -3d + 13. / Your problem should look like: d = -3d + 13
Fourth, add 3d to both sides. / Your problem should look like: d + 3d = 13
Fifth, add d + 3d to get 4d. / Your problem should look like: 4d = 13
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Answer as fraction: d =  \frac{13}{4}
Answer as decimal: d = 3.25

3 0
3 years ago
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