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Jet001 [13]
4 years ago
14

Write g(n)=2n(3-n)+5 in standard form

Mathematics
1 answer:
wolverine [178]4 years ago
6 0
Standard form is an²+bn+c

expand
distribute
2n(3-n)+5=
2n(3)+2n(-n)+5=
6n-2n²+5=
-2n²+6n+5


g(n)=-2n²+6n+5 is standard form
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Read 2 more answers
A counselor records the number of disagreements (per session) among couples during group counseling sessions. If the number of d
Shalnov [3]

Answer:

P(X>4)=P(\frac{X-\mu}{\sigma}>\frac{4-\mu}{\sigma})=P(Z>\frac{4-4.4}{0.4})=P(z>-1)

And we can find this probability with the complement rule:

P(z>-1)=1-P(z

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

Solution to the problem

Let X the random variable that represent the disagreements of a population, and for this case we know the distribution for X is given by:

X \sim N(4.4,0.4)  

Where \mu=4.4 and \sigma=0.4

We are interested on this probability

P(X>4)

And the best way to solve this problem is using the normal standard distribution and the z score given by:

z=\frac{x-\mu}{\sigma}

If we apply this formula to our probability we got this:

P(X>4)=P(\frac{X-\mu}{\sigma}>\frac{4-\mu}{\sigma})=P(Z>\frac{4-4.4}{0.4})=P(z>-1)

And we can find this probability with the complement rule:

P(z>-1)=1-P(z

3 0
3 years ago
I need help on number 5 please and thanks
aleksandrvk [35]

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John paid 18% of his earnings on entertainment.

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3 years ago
let a = (a1, a2) and b = (b1, b2) and c = (c1,c2) be three non zero vectors. if a1b2 - a2b1 is not equal to 0. then show three a
Ksenya-84 [330]

Consider the contrapositive of the statement you want to prove.

The contrapositive of the logical statement

<em>p</em> ⇒ <em>q</em>

is

¬<em>q</em> ⇒ ¬<em>p</em>

In this case, the contrapositive claims that

"If there are no scalars <em>α</em> and <em>β</em> such that <em>c</em> = <em>α</em><em>a</em> + <em>β</em><em>b</em>, then <em>a₁b₂</em> - <em>a₂b₁</em> = 0."

The first equation is captured by a system of linear equations,

\begin{cases}c_1 = \alpha a_1 + \beta b_1\\ c_2 = \alpha a_2 + \beta b_2\end{cases}

or in matrix form,

\begin{pmatrix}c_1\\c_2\end{pmatrix} = \begin{pmatrix}a_1&b_1\\a_2&b_2\end{pmatrix}\begin{pmatrix}\alpha\\\beta\end{pmatrix}

If this system has no solution, then the coefficient matrix on the right side must be singular and its determinant would be

\begin{vmatrix}a_1&b_1\\a_2&b_2\end{vmatrix} = a_1b_2-a_2b_1 = 0

and this is what we wanted to prove. QED

3 0
3 years ago
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