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lianna [129]
2 years ago
9

SAT math scores have a bell-shaped distribution with a mean of 515 and a standard deviation of 114.

Mathematics
1 answer:
nadezda [96]2 years ago
3 0

Hey there mate :)

Analysis :

401 = 515 - 114 = u - 6

And

173 = 515 - 3.114 = u - 36

=  \frac{p(u - 36 < u < u + 36) - p(u - 6 < u < u + 6)}{2}

=  \frac{99.73percent - 68.21percent}{2}

=  \frac{31.46percent}{2}

= {15.73percent}

Then, calculating,

u - 26 = 515 - 2 \times 114 = 287

And

u + 26 = 515 + 2 \times 114 = 743

Then,

The \: Range = (287,743)

Final Answers :-

<u>1</u><u>5</u><u>.</u><u>7</u><u>3</u><u> </u><u>%</u>

<u>(</u><u>2</u><u>8</u><u>7</u><u>,</u><u>7</u><u>4</u><u>3</u><u>)</u>

~Benjemin360

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Step-by-step explanation:

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4. Do the same with the numbers above the median.

5. Take the medians of the lower numbers and higher and find the difference,

6. Thats your interquartile range

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Read 2 more answers
Air-USA has a policy of booking as many as 24 persons on an airplane that can seat only 22. (Past studies have revealed that onl
Alex17521 [72]

Answer:

B. no, it is not low enough

A. no, it is not low enough

Step-by-step explanation:

Given that Air-USA has a policy of booking as many as 24 persons on an airplane that can seat only 22.

Prob for  a random person booked arrive for flight = 0.86

No of persons who books and arrive for flight, X is binomial, since there are two outcomes and each person is independent of the other

The probability that if Air-USA books 24 persons, not enough seats will be available

= P(X=23)+P(x=24)

= 0.1315

B. no, it is not low enough

-------------------------------

The prob we got is >10% also

A. no, it is not low enough

7 0
3 years ago
Two isosceles triangles share the same base. Prove that the medians to this base are collinear. (There are two cases in this pro
Anna71 [15]

Answer:

Given: Two Isosceles triangle ABC and Δ PBC having same base BC. AD is the median of Δ ABC and PD is the median of Δ PBC.

To prove: Point A,D,P are collinear.

Proof:

→Case 1.  When vertices A and P are opposite side of Base BC.

In Δ ABD and Δ ACD

AB= AC   [Given]

AD is common.

BD=DC  [median of a triangle divides the side in two equal parts]

Δ ABD ≅Δ ACD [SSS]

∠1=∠2 [CPCT].........................(1)

Similarly, Δ PBD ≅ Δ PCD [By SSS]

 ∠ 3 = ∠4 [CPCT].................(2)

But,  ∠1+∠2+ ∠ 3 + ∠4 =360° [At a point angle formed is 360°]

2 ∠2 + 2∠ 4=360° [using (1) and (2)]

∠2 + ∠ 4=180°

But ,∠2 and ∠ 4 forms a linear pair i.e Point D is common point of intersection of median AD and PD of ΔABC and ΔPBC respectively.

So, point A, D, P lies on a line.

CASE 2.

When ΔABC and ΔPBC lie on same side of Base BC.

In ΔPBD and ΔPCD

PB=PC[given]

PD is common.

BD =DC [Median of a triangle divides the side in two equal parts]

ΔPBD ≅ ΔPCD  [SSS]

∠PDB=∠PDC [CPCT]

Similarly, By proving ΔADB≅ΔADC we will get,  ∠ADB=∠ADC[CPCT]

As PD and AD are medians to same base BC of ΔPBC and ΔABC.

∴ P,A,D lie on a line i.e they are collinear.




3 0
3 years ago
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