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4 years ago

16×16×16 equivalent to?

Mathematics

2 answers:

navik [9.2K]4 years ago

5 0

Answer:

16³= 4096

Step-by-step explanation:

cluponka [151]4 years ago

4 0

$16\cdot16\cdot16 = 16^3 = (2^4)^3 = 2^{12} = 4\cdot 2^{10} = 4096$

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The drama club is selling candles for a fundraiser. They spend $100 on the candles and sell them for $4.50 each. The inequality

atroni [7]

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3 years ago

A 4-column table with 3 rows. The first column has no label with entries less than 20 square miles, greater than 20 square miles

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2 years ago

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The mean number of automobiles entering a mountain tunnel per two-minute period is one. An excessive number of cars entering the

viva [34]

Answer:

A Poisson model seems reasonable for this problem, since we have the mean during the time interval.

There is a 1.9% probability that the number of autos entering the tunnel during a two-minute period exceeds three.

Step-by-step explanation:

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

$P(X = x)=\frac{e^{-\mu}*\mu^{x}}{(x)!}$

In which

x is the number of sucesses

e = 2.71828 is the Euler number

$\mu$ is the mean in the given time interval.

The mean number of automobiles entering a mountain tunnel per two-minute period is one.

This means that $\mu = 1$ .

For a Poisson model to be reasonable, we only need the mean during the time interval. So yes, a Poisson model seems reasonable for this problem.

Find the probability that the number of autos entering the tunnel during a two-minute period exceeds three.

We want to find $P(X>3)$

Either this number is less or equal to 3, or it exceeds 3. The sum of the probabilities is decimal 1. So:

$P(X \leq 3) + P(X > 3) = 1$

$P(X > 3) = 1 - P(X \leq 3)$

In which

$P(X \leq 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)$

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

$P(X = 0) = \frac{e^{-1}*1^{0}}{(0)!} = 0.3679$

$P(X = 1) = \frac{e^{-1}*1^{1}}{(1)!} = 0.3679$

$P(X = 2) = \frac{e^{-1}*1^{2}}{(2)!} = 0.1839$

$P(X = 3) = \frac{e^{-1}*1^{3}}{(3)!} = 0.0613$

$P(X \leq 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) = 0.3679 + 0.3679 + 0.1839 + 0.0613 = 0.981$

Finally

$P(X > 3) = 1 - P(X \leq 3) = 1 - 0.981 = 0.019$

There is a 1.9% probability that the number of autos entering the tunnel during a two-minute period exceeds three.

8 0

4 years ago

The interquartile range of the data set is 4. 2, 2, 3, 3, 4, 5, 5, 6, 7, 9, 12

Rzqust [24]

Answer:

The IQR or the interquartile range of the following data set would be 4.

Step-by-step explanation:

2, 2, 3, 3, 4, 5, 5, 6, 7, 9, 12

Median: 5

Lower quartile: 3

Upper quartile: 7

Interquartile range: 7 - 3 = 4

7 0

4 years ago

5y - 25x =10 I need someone to help me solve this asap

ruslelena [56]

Answer:slope=10.000/2.000=5.000

x-intercept=2/5=minus 0.40000

y-intercept=2/1=2.00000

Step-by-step explanation:

8 0

4 years ago