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2 years ago

Brianna is making fruit dip. The

Mathematics

1 answer:

STALIN [3.7K]2 years ago

6 0

The answer is 6 servings

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-3(5)^2-(-2)(-8) simplify

avanturin [10]

To do this, we're going to use the order of operations (PEMDAS):

P - Parentheses

E - Exponents

M - Multiplication

D - Division

A - Addition

S - Subtraction

First let's do parentheses, there isn't anythig in parentheses we need to simplify, so we can skip this step.

Next let's look for exponents. I see we have a $5^2$ so let's replace that with $5^2=25$ :

$-3(25)-(-2)(-8)$

Now let's look for multiplcation. We know that things that are right next to eachother in parentheses represent multiplcation, so let's simply this more:

$-3(25)=-75$

$(-2)(-8)=16$

$-75-16$

And now we're left with a simple problem we know how to solve.

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2 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

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2 years ago