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3 years ago

Simplify the following expression five to the power of -3

Mathematics

1 answer:

skelet666 [1.2K]3 years ago

3 0

Hmm I wonder too this is my question exactly crazy.

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Which solid figures have their volumes calculated using the formula V = One-thirdBh, where B is the area of the base and h is th

eimsori [14]

Answer:

A. Right Cone

C. Hexagonal prism

Step-by-step explanation:

The volume of a cone with height h and radius r is 1/3πr^2h, which is exactly one third the volume of the smallest cylinder that it fits inside.

The volume of a hexagonal pyramid with height h and radius r is 1/3πr^2h, which is exactly one third the volume of the smallest rectangular prism that it fits inside.

4 0

3 years ago

Read 2 more answers

Jason has a lemonade stand but he already made $26 he charges his customers one dollar per lemonade Kaylee has a lemonade stand

Aleksandr-060686 [28]

Answer: B. X+26<2x+16

Step-by-step explanation:

Let x be the number of lemonade Kaylee needs to sell to make a higher profit than Jason.

Given: Jason has made $26. he charge $1 per lemonade.

Total earnings after x lemonades = $26+($1) x

= $(26+x)

Kaylee has made $16. he charge $2 per lemonade.

Total earnings after x lemonades = $16+($2) x

= $(16+2x)

Required inequality : (16+2x)> (26+x)

i.e. correct option is B.

6 0

3 years ago

Airline passengers arrive randomly and independently at the passenger-screening facility at a major international airport. The m

marshall27 [118]

Answer:

1. 0.0000454

2. 0.01034

3. 0.0821

4. 0.918

Step-by-step explanation:

Let X be the random variable denoting the number of passengers arriving in a minute. Since the mean arrival rate is given to be 10,

$X \sim Poi(\lambda = 10)$

1. Requires us to compute

$P(X = 0) = e^{-10} \frac{10^0}{0!} = 0.0000454$

2. We need to compute $P(X \leq 3) = P(X =0) + P(X =1) + P(X =2) + P(X =3)$

$P(X =1) = e^{-10} \frac{10^1}{1!} = 0.000454$

$P(X =2) = e^{-10} \frac{10^2}{2!} = 0.00227$

$P(X =3) = e^{-10} \frac{10^3}{3!} = 0.00757$

$P(X \leq 3) =0.0000454+ 0.000454 + 0.00227 + 0.00757 = 0.01034$

3. The expected no. of arrivals in a 15 second period is = $10 \times \frac{1}{4} = 2.5$ . So if Y be the random variable denoting number of passengers arriving in 15 seconds,

$Y \sim Poi(2.5)$

$P(Y=0) = e^{-2.5} \frac{2.5^0}{0!} = 0.0821$

4. Here we use the fact that Y can take values $0,1, \dotsc$ . So, the event that "Y is either 0 or $\geq$ 1" is a sure event ( i.e it has probability 1 ).