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3 years ago

Write each of the numbers 1,8,27,64, and 125 as a based raised to the third power

Mathematics

1 answer:

KatRina [158]3 years ago

6 0

$1 = {1}^{3 } \\ 8 = {2}^{3} \\ 27 = {3}^{3} \\ 64 = {4}^{3}$
$125 = {5}^{3}$

good luck

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(a) A parachutist lands at a point on the line between the points A and B, and the target is an operation at A. The operation fa

mr Goodwill [35]

Answer:

a) $\frac{B-\frac{A+5B}{6}}{B-A}= \frac{6B -A-5B}{6(B-A)}=\frac{B-A}{6(B-A)}=\frac{1}{6}$

b) $P(X>1) = 1-P(X\leq 1)=1-\int_{0}^1 \frac{1^{2} x^{2-1} e^{- x}}{\gamma(2)}=0.736$

Step-by-step explanation:

Part a

We assume that the parachutist lands at random point in the interval (x=A,y=B) we have a continuous random variable X. And the distribution of X would be uniform $Y\sim Unif(A,B)$ . And the density function would be given by:

$f(x) =\frac{1}{B-A} , A$

And 0 for other case.

The operation fails, if parachutist's distance to A is more than five times as much as her distance to B.

So the point P in the interval (A,B) at which the distance to A is exactly 5 times the distance to B is given by:

$P= A + \frac{5}{6} (B-A)= \frac{6A +5B -5A}{6}=\frac{A+5B}{6}$

And we can find the probability desired like this:

$P(d(P,A) \geq 5 d(P,B))= P(\frac{A+5B}{6} < X< B)$

And from the cumulative distribution function of X ficen by $F(X)\frac{X-A}{B-A}$ we got:

$\frac{B-\frac{A+5B}{6}}{B-A}= \frac{6B -A-5B}{6(B-A)}=\frac{B-A}{6(B-A)}=\frac{1}{6}$

Part b

For this case we assume that $X\sim Gamma (2,1)$

On this case we assume that $\alpha=2, \beta= 1$

The density function for the Gamma distribution is given by:

$P(X)= \frac{\beta^{\alpha} x^{\alpha-1} e^{-\beta x}}{\gamma(\alpha)}$

And on this case we can find the probability using the complement rule like this:

$P(X>1) = 1-P(X\leq 1)=0.736$

We can solve this problem with the following excel code:

"=1-GAMMA.DIST(1;2;1;TRUE)"

And if we do it by hand we need to do this:

$P(X>1) = 1-P(X\leq 1)=1-\int_{0}^1 \frac{1^{2} x^{2-1} e^{- x}}{\gamma(2)}=0.736$

6 0

3 years ago

A man whose height is 2.1 m casts a shadow that is 3.2 m in length. What is the approximate height of a building that casts a sh

Colt1911 [192]

Answer:

ok so first

2.1 divided by 3.2 =0.65625

0.65625*11.5=7.546875

So the answer is B

Hope This Helps!!!

6 0

3 years ago

The gardener is going to cover the path with sand. If one bag of sand can cover 5yd^2, how many bags must be a whole number.

Nesterboy [21]

5 a bag of bags be can cover2 garden

5 0

3 years ago

I need help with this question

nikdorinn [45]

The area of each triangular face is
.. A = (1/2)*b*h
.. Aside = (1/2)*(6 cm)*(5 cm) = 15 cm^2

The area of the triangular base is
.. Abase = ((√3)/4)*s^2 = (√3)/4*(6 cm)^2 = 9√3 cm^2

The total surface area is the base area plus the area of the three sides.
.. Atotal = Abase + 3*Aside
.. = 9√3 cm^2 +3*15 cm^2
.. = (45 +9√3) cm^2
.. ≈ 60.6 cm^2

Actually, there's an error in the picture. The height of the side should be 4, and the length of the edge should be 5. Making that adjustment, the total area is 51.6 cm^2.

It is hard to tell what is intended. Not all answers are showing, so we can't "reverse-engineer" the problem from the answers.

7 0

3 years ago

Thirty car pass through a certain toll booth each minute.at that rate,how many cars are expected to pass through the toll booth

BigorU [14]

D. 30 Cars p/m x 30 min = 900

8 0

3 years ago

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