One card is drawn at random from a deck of cards. What is the probability that the card

The volume of a cone is 62.8 cubic inches. The height of the cone is 15 inches. What is the radius of the cone, rounded to the n

Setler79 [48]

R = sqrt 3 * (V /( pi * h))
V = 62.8
pi = 3.14
h = 15
now we sub
R = sqrt 3 * (62.8 / (3.14 * 15)
R = sqrt 3 * (62.6 / 47.1)
R = sqrt 3 * 1.33
R = sqrt 3.99
R = 1.9 rounds to 2 inches <===

3 0

3 years ago

Read 2 more answers

Solve for m, what is k=mv squared/2

scoray [572]

Answer:

Solve for

K

by simplifying both sides of the equation, then isolati.ng the variable.

K

=

m

v

2 i.f yo.u wa.nt th.e re.al answ.er g.o h.ere: >>>>https://www.math.way.com/po.pular-prob.lems/Alg.ebra/229798

2

Step-by-step explanation:

4 0

3 years ago

Matteo spends a total of 38 min exercising. He walks for 6 min to warm up and then runs at a constant rate of 8 min per mile for

umka2103 [35]

Answer:

No, he not correct because he only ran 4 mi

Step-by-step explanation:

If he walks for 6 min out of the 38 min, then he ran for 32 min.

d = rt 1 mi = 8r

r = 1/8 = 1 mi/8 min Since he ran 32 min we have

d = 1/8(32) = 4 mi.

No, he not correct because he only ran 4 mi

6 0

3 years ago

Let X denote the distance (m) that an animal moves from its birth site to the first territorial vacancy it encounters. Suppose t

andrezito [222]

Answer and Step-by-step explanation: For an exponential distribution, the probability distribution function is:

f(x) = λ. $e^{-\lambda.x}$

and the cumulative distribution function, which describes the probability distribution of a random variable X, is:

F(x) = 1 - $e^{-\lambda.x}$

(a) Probability of distance at most 100m, with λ = 0.0143:

F(100) = 1 - $e^{-0.0143.100}$

F(100) = 0.76

Probability of distance at most 200:

F(200) = 1 - $e^{-0.0143.200}$

F(200) = 0.94

Probability of distance between 100 and 200:

F(100≤X≤200) = F(200) - F(100)

F(100≤X≤200) = 0.94 - 0.76

F(100≤X≤200) = 0.18

(b) The mean, E(X), of a probability distribution is calculated by:

E(X) = $\frac{1}{\lambda}$

E(X) = $\frac{1}{0.0143}$

E(X) = 69.93

The standard deviation is the square root of variance,V(X), which is calculated by:

σ = $\sqrt{\frac{1}{\lambda^{2}} }$

σ = $\sqrt{\frac{1}{0.0143^{2}} }$

σ = 69.93

Distance exceeds the mean distance by more than 2σ:

P(X > 69.93+2.69.93) = P(X > 209.79)

P(X > 209.79) = 1 - P(X≤209.79)

P(X > 209.79) = 1 - F(209.79)

P(X > 209.79) = 1 - (1 - $e^{-0.0143*209.79}$ )

P(X > 209.79) = 0.0503

(c) Median is a point that divides the value in half. For a probability distribution:

P(X≤m) = 0.5

$\int\limits^m_0 f({x}) \, dx$ = 0.5

$\int\limits^m_0 {\lambda.e^{-\lambda.x}} \, dx$ = 0.5

$\lambda.\frac{e^{-\lambda.x}}{-\lambda}$ = $-e^{-\lambda.x} + e^{0}$

$1 - e^{-\lambda.m}$ = 0.5

$-e^{-\lambda.m}$ = - 0.5

ln( $e^{-0.0143.m}$ ) = ln(0.5)

-0.0143.m = - 0.0693

m = 48.46

6 0

4 years ago

Tell me the answers please!

il63 [147K]

18 is D and 19 is C you're welcome

4 0

4 years ago