A right cylinder picture to the right its base radius is 5 cm and its height is 16 cm. all in the pic if possible

Mathematics

2 answers:

Oduvanchick [21]2 years ago

8 0

Answer:

1.) 1256.24

Step-by-step explanation:

Too blurry...

search up cylinder volume on GoogIe

it'll help.

$\pi r^{3} h$ is formula for volume of cylinder

SashulF [63]2 years ago

5 0

Answer:

$\small\mathfrak\pink{s} \small\mathfrak\purple{o} \small\mathfrak\blue{l} \small\mathfrak\red{u}\small\mathfrak\green{t} \small\mathfrak\orange{i}\small\mathfrak\blue{o} \small\mathfrak\red{n:-}$

157 cm²
502.4 cm²
659.4cm²

 $\\ \\$ 

Hope it helps

~ʆᵒŕ∂ཇꜱꜹⱽẻⱮë

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Suppose that the time (in hours) required to repair a machine is an exponentially distributed random variable with mean 1/0.41/0

lbvjy [14]

Answer:

a) P(t>3)=0.30

b) P(t>10|t>9)=0.67

Step-by-step explanation:

We have a repair time modeled as an exponentially random variable, with mean 1/0.4=2.5 hours.

The parameter λ of the exponential distribution is the inverse of the mean, so its λ=0.4 h^-1.

The probabity that a repair time exceeds k hours can be written as:

$P(t>k)=e^{-\lambda t }=e^{-0.4t}$

(a) the probability that a repair time exceeds 3 hours?

$P(t>3)=e^{-0.4*3}=e^{-1.2}= 0.30$

(b) the conditional probability that a repair takes at least 10 hours, given that it takes more than 9 hours?

The exponential distribution has a memoryless property, in which the probabilities of future events are not dependant of past events.

In this case, the conditional probability that a repair takes at least 10 hours, given that it takes more than 9 hours is equal to the probability that a repair takes at least (10-9)=1 hour.

$P(t>10|t>9)=P(t>1)$