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monitta
3 years ago
8

Suppose 12% of students are veterans. From a sample of 263 students, how unusual would it be to have less than 22 veterans?

Mathematics
1 answer:
kkurt [141]3 years ago
7 0

Answer:

Yes, both np and n(1-p) are ≥ 10

Mean = 0.12 ; Standard deviation = 0.02004

Yes. There is a less than 5% chance of this happening by random variation. 0.034839

Step-by-step explanation:

Given that :

p = 12% = 0.12 ;

Sample size, n = 263

np = 263 * 0.12 = 31.56

n(1 - p) = 263(1 - 0.12) = 263 * 0.88 = 231.44

According to the central limit theorem, distribution of sample proportion approximately follow normal distribution with mean of p = 0.12 and standard deviation sqrt(p*(1 - p)/n) = sqrt (0.12 *0.88)/n = sqrt(0.0004015) = 0.02004

Z = (x - mean) / standard deviation

x = 22 / 263 = 0.08365

Z = (0.08365 - 0.12) / 0.02004

Z = −1.813872

Z = - 1.814

P(Z < −1.814) = 0.034839 (Z probability calculator)

Yes, it is unusual

0.034 < 0.05 (Hence, There is a less than 5% chance of this happening by random variation.

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saul85 [17]

\large\underline{\sf{Solution-}}

<u>Let us assume that:</u>

\sf \longmapsto x =  \sqrt{6 +  \sqrt{6 +  \sqrt{6 + ... \infty } } }

We can also write it as:

\sf \longmapsto x =  \sqrt{6 +  x }

Squaring both sides, we get:

\sf \longmapsto  {x}^{2}  =6 +  x

\sf \longmapsto  {x}^{2} - x - 6  =0

By splitting the middle term:

\sf \longmapsto  {x}^{2} - 3x + 2x - 6  =0

\sf \longmapsto x(x - 3) + 2(x - 3 ) =0

\sf \longmapsto (x+ 2)(x - 3 ) =0

<u>Therefore:</u>

\longmapsto\begin{cases} \sf (x+ 2) =0 \\ \sf (x - 3) = 0 \end{cases}

\sf \longmapsto x =  - 2,3

<u>But x cannot be negative. </u>

\sf \longmapsto x = 3

Therefore, the value of the expression is 3.

\large\underline{\sf{Verification-}}

Given:

\sf\longmapsto x=3

We can also write it as:

\sf\longmapsto x = \sqrt{9}

\sf\longmapsto x = \sqrt{6+3}

\sf\longmapsto x = \sqrt{6 + \sqrt{9}}

\sf\longmapsto x = \sqrt{6 + \sqrt{6+3}}

\sf\longmapsto x = \sqrt{6 + \sqrt{6+\sqrt{9}}}

\sf\longmapsto x = \sqrt{6 + \sqrt{6+\sqrt{6+3}}}

This pattern will continue.

\sf\longmapsto x = \sqrt{6 + \sqrt{6+\sqrt{6+...\infty}}}

7 0
3 years ago
how do you solve a system of equations approximately given tables without using equations or graphing?
solong [7]

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If you only have tables, this means that you need to have one table for each equation:

For example, if you are working only with two variables, x and y, in those tables you can see the pints (x, y) that belong to each equation.

Now, a point (x, y) will be a solution of the system of equations only if it belongs to the data table for each equation

This would mean that if you graph those data sets, the graphs will intersect at the point (x, y) that belongs to all the tables of data.

another way may be using the data in the tables to construct the equations, but you said that you only want to use the tables, so this method can be discarded.

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3 years ago
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Answer:

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Step-by-step explanation:

6 0
3 years ago
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Lemur [1.5K]

Answer:

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Step-by-step explanation:

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α = 55°

o: opposite  = 2.0m

h: hypotenuse

sin α = o/h

cos α= a/h

tan α = o/a

we see that it has (angle, hypotenuse, opposite)

we look at which meets those data between the sine, cosine and tangent

is the sine

sin α = o/ah

Now we replace the values ​​and solve

sin 55 = 2.0/h

0.81915 = 2.0/h

h = 2.0 / 0.81915

h = 2.4415 m

round to the nearest tenth

h = 2.4415 = 2.4 m

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Answer:

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