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3 years ago

Anything to the power of 0 is always 1 right??

Mathematics

2 answers:

lana66690 [7]3 years ago

6 0

Yes, that is correct.

Shtirlitz [24]3 years ago

4 0

Yes anything to the power of 0 is one in the contrary anything to the power of 1 its it self

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40 POINTS PLEASE HELP AND HURRY

pentagon [3]

Answer:

The histogram that represents the correct frequency of each class interval in the given data set, is: histogram B.

What is a Histogram?

A histogram is a type of bar graph that uses bins to represent the frequency of a class interval.

Given the data points above, the histogram in option B has the correct bars that represent the appropriate frequency of each class interval.

Therefore, the histogram that represents the data set given is Histogram B.

Learn more about histograms on:

brainly.com/question/18198700

Please Do this:

<h2><u><em>Please give me brainleist!</em></u></h2>

3 0

2 years ago

What is the best answer to solution of x^2 + 16 = 0?

kolezko [41]

Answer:

4. C. x = 4 or -4

Step-by-step explanation:

since both terms are perfect squares, factor using the difference of squares formula, a² - b² = ( a + b ) ( $\alpha$ - b ) where

5 0

3 years ago

Consider the sequence:

Vera_Pavlovna [14]

Answer:

a_n = 2^n + 3

Step-by-step explanation:

The first differences have a geometric progression, so the explicit definition will be an exponential function. (It cannot be modeled by a linear or quadratic function.) The above answer is the only choice that is an exponential function.

First differences are ...

(7-5=)2, 4, 8, 16

8 0

3 years ago

Read 2 more answers

Find the domain of the relation:

Lerok [7]

The domain is -3, -1, 0, 1, 5

8 0

3 years ago

Assume that a policyholder is four times more likely to file exactly two claims as to file exactly three claims. Assume also tha

kykrilka [37]

Answer:

1.3125

Step-by-step explanation:

Given that our random variable $X$ follows a Poisson distribution $P(X=k)=\frac{\lambda^k e^-^p}{k!} \ \ \ \ \ \ \ \ p=\lambda$

Evaluate the formula at $k=2,3:$

$P(X=2)=0.5\lambda^2e^{-\lambda}\\\\P(X=3)=\frac{1}{6}\lambda^2e^{-\lambda}$

#since

$4P(X=2)=P(X=3);\\\\0.5\lambda^2e^{-\lambda}=4\times\frac{1}{6}\lambda^3e^{-\lambda}\\\\0.5\lambda^2=\frac{2}{3}\lambda^3\\\\0.75=\lambda$

The mean and variance of the Poisson distributed random variable is equal to $\lambda$ :

$\mu=\lambda=0.75\\\sigma ^2=\lambda=0.75$

#By property variance:

$\sigma ^2=V(X)=E(X^2)-(E(X))^2=E(X^2)\\\\E(X^2)=\sigma^2+\mu^2=0.75+0.75^2=1.3125$

The expectation is 1.3125

4 0

3 years ago