Write the word sentence as an inequality.<br><br> One plus a number y<br> is no more than −13

Delicious77 [7]

Answer:

$n=-2,\:n=-1$

Step-by-step explanation:

$\left(n+2\right)\left(n+1\right)=0\\\mathrm{Using\:the\:Zero\:Factor\:Principle:\quad \:If}\:ab=0\:\mathrm{then}\:a=0\:\mathrm{or}\:b=0\:\left(\mathrm{or\:both}\:a=0\:\mathrm{and}\:b=0\right)\\\mathrm{Solve\:}\:n+2=0:\quad n=-2\\n+2=0\\\mathrm{Subtract\:}2\mathrm{\:from\:both\:sides}\\n+2-2=0-2\\\mathrm{Simplify}\\n=-2\\\mathrm{Solve\:}\:n+1=0:\quad n=-1\\n+1=0\\\mathrm{Subtract\:}1\mathrm{\:from\:both\:sides}\\n+1-1=0-1\\Simplify\\n=-1\\$ $\mathrm{The\:solutions\:to\:the\:quadratic\:equation\:are:}\\n=-2,\:n=-1$

7 0

3 years ago

Read 2 more answers

5x-6=29. What does x equal

anastassius [24]

Answer:

Solution given:

5x-6=29.

5x=29+6

x=35/5=7

x=7 is your answer

7 0

2 years ago

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Omar deposited $4000 into an account with 4.4% interest, compounded quarterly. Assuming that no withdrawals are made, how much w

melamori03 [73]

The formula you want is: fv=p(1+int/c)^(nc)
future value
principal
int
compound
years

7 0

1 year ago

What is the mean of the set of numbers?

Llana [10]

The mean of your question would be -2.5

3 0

2 years ago

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Determine whether a probability distribution is given. If a probability distribution is given, find its mean and standard deviat

drek231 [11]

Answer:

$E(X) = \sum_{i=1}^n X_i P(X_i) = 0*0.031 +1*0.156+ 2*0.313+3*0.313+ 4*0.156+ 5*0.031 = 2.5$

We can find the second moment given by:

$E(X^2) = \sum_{i=1}^n X^2_i P(X_i) = 0^2*0.031 +1^2*0.156+ 2^2*0.313+3^2*0.313+ 4^2*0.156+ 5^2*0.031 =7.496$

And we can calculate the variance with this formula:

$Var(X) =E(X^2) -[E(X)]^2 = 7.496 -(2.5)^2 = 1.246$

And the deviation is:

$Sd(X) = \sqrt{1.246}= 1.116$

Step-by-step explanation:

For this case we have the following probability distribution given:

X 0 1 2 3 4 5

P(X) 0.031 0.156 0.313 0.313 0.156 0.031

The expected value of a random variable X is the n-th moment about zero of a probability density function f(x) if X is continuous, or the weighted average for a discrete probability distribution, if X is discrete.

The variance of a random variable X represent the spread of the possible values of the variable. The variance of X is written as Var(X).

We can verify that:

$\sum_{i=1}^n P(X_i) = 1$

And $P(X_i) \geq 0, \forall x_i$

So then we have a probability distribution

We can calculate the expected value with the following formula:

$E(X) = \sum_{i=1}^n X_i P(X_i) = 0*0.031 +1*0.156+ 2*0.313+3*0.313+ 4*0.156+ 5*0.031 = 2.5$

We can find the second moment given by:

$E(X^2) = \sum_{i=1}^n X^2_i P(X_i) = 0^2*0.031 +1^2*0.156+ 2^2*0.313+3^2*0.313+ 4^2*0.156+ 5^2*0.031 =7.496$

And we can calculate the variance with this formula:

$Var(X) =E(X^2) -[E(X)]^2 = 7.496 -(2.5)^2 = 1.246$

And the deviation is:

$Sd(X) = \sqrt{1.246}= 1.116$

6 0

3 years ago

Write the word sentence as an inequality. One plus a number y is no more than −13