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3 years ago

How should you classify a closed shape with 5 congruent side lengths?

2 answers:

Elena-2011 [213]3 years ago

7 0

The correct answer is a Regular Pentagon, but it could also be a pentagon, because a regular pentagon is technically a pentagon.

UNO [17]3 years ago

6 0

It would be regular, in this case a regular polygon or pentagon.

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Which is the simplified form for 10 -5x<20

sergij07 [2.7K]

X > -2 is the answer

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4 years ago

F(x)=2(x)^2+5 sqrt (x+2)<br><br> f(0)

krek1111 [17]

F(0) = replace every x with 0

$2 ({0})^{2} + 5 \sqrt{0 + 2} \\ = 2 + 5 \sqrt{2}$
=9.0710678119
=9.07

7 0

3 years ago

2(4y-5)=70<br><br>how u do this

steposvetlana [31]

Answer:

Step-by-step explanation:

First you distribute so two times 4Y is 6Y then you multiply the two in the five which gives you 2Y -5 equals 70

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4 years ago

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The city of Joyville has 87,388 residents round the number of people who live in joyville

Yuliya22 [10]

Answer:

there is no answer

but ok

Step-by-step explanation:

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3 years ago

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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

4 years ago