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icang [17]
3 years ago
12

The same sample of 25 seniors from the urban school district with the mean and standard deviation N(450, 100). A 95% confidence

interval for µ for the population of seniors with a margin of error of ± 25 is used. What is the smallest sample size we can take to achieve this same margin of error?
Mathematics
1 answer:
kvv77 [185]3 years ago
3 0
The margin of error of a sample is given by:
B = z(α/2) σ/√n; where B is the margin of error = 25, z(α/2) is the level of significant = 1.96, σ is the standard deviation = 100 and n is the sample size.

25 = 1.96(100/√n)
100/√n = 25/1.96 = 12.76
√n = 100/12.76 = 7.84
n = (7.84)^2 = 61.47

Therefore, the smallest sample size required is 62 seniors.
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12. In the given figure, RS is parallel to PQ, If RS = 3 cm, PQ = 6 cm and ar(∆TRS) = 15cm³, then ar (∆TPQ) = ? (a) 70 cm² (b) 5
Gnesinka [82]

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

Given that,

In <u>triangle TPQ, </u>

  • RS || PQ,

  • RS = 3 cm,

  • PQ = 6 cm,

  • ar(∆ TRS) = 15 sq. cm

As it is given that, <u>RS || PQ</u>

So, it means

⇛∠TRS = ∠TPQ [ Corresponding angles ]

⇛ ∠TSR = ∠TPQ [ Corresponding angles ]

\rm\implies \: \triangle TPQ \:  \sim \: \triangle TRS \:  \:  \:  \:  \:  \:  \{AA \}

<u>Now, We know </u>

Area Ratio Theorem,

This theorem states that :- The ratio of the area of two similar triangles is equal to the ratio of the squares of corresponding sides.

\rm\implies \:\dfrac{ar( \triangle \: TPQ)}{ar( \triangle \: TRS)}  = \dfrac{ {PQ}^{2} }{ {RS}^{2} }

\rm\implies \:\dfrac{ar( \triangle \: TPQ)}{15}  = \dfrac{ {6}^{2} }{ {3}^{2} }

\rm\implies \:\dfrac{ar( \triangle \: TPQ)}{15}  = \dfrac{36 }{9}

\rm\implies \:\dfrac{ar( \triangle \: TPQ)}{15}  = 4

\rm\implies \:ar( \triangle \: TPQ)  = 60 \:  {cm}^{2}

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Step-by-step explanation:

I think the problem was.  

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Given

\begin{gathered} x^2=100 \\ x=10 \\ x=-10 \end{gathered}

Procedure

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