A teacher informs her cell biology class (of 500+ students) that a test was very difficult, but the grades would be curved. Scores on the test were normally distributed with a mean of 37 and a standard deviation of 5. The maximum possible score on the test was 100 points. Because of partial credit, scores were recorded with 1 decimal point accuracy. (Thus, a student could earn a 37.6, but not a 36.77.) The grades are curved according to the following scheme. Find the numerical limits for each letter grade. Letter Scheme Interval A Top 15% B to C D F Scores above the bottom 74% and below the top 15% Scores above the bottom 26% and below the top 26% Scores above the bottom 15% and below the top 74% Bottom 15%

Q: (a)

A: Given,           jj

Q: dx dy 26.

A:

Q: 6. dy = 3 — 2у

A:

Q: fr рор pår (1+ (1 + ln r)

A:

Q: from e'dv Ve4v -16

A:

Q: 1 dx 2 x"Vx-3 Evaluate

A:

Q: 1 -dx V 9x² + 18x + 45 2

A:

Q: 2 X- - SS. T dydx

A: ∫1-1∫0x2-1dy dxWe need to evaluate the above double integral.

Q: 4 dx 8)

A: Consider the given definite integral.

Q: S sec ydy

A:

Q: S csc? (2y)/(9 + cot²(2y)) dy

A:

Q: (3 In z + 2)4 dz Z

A: The given integral is (3 lnz +2)4/z.

Q: 2. 2 PCn):芝(-リ 2h %3D j = 1

A:

Q: *43メー バ+5 K

A: Please see the answer below.

Q: 25 2) V1x-91

A: To evaluate first: ∫dx|x-9| Assume the positive factors and multiply the correction factors:…

Q: .2 dx r6 +1

A:

Q: Eval the Pt: dx

A:

Q: dx 3— 2л — т2

A:

Q: dx 15 + 2x – x²

A: ∫dx15+2x-x2

Q: dx -(r+

A: We have to evaluate the following integral: ∫dx4-x+22

Q: fnd the snh. Jnt. 6x+3 da

A: Find the integration

Q: 7) (3x* – 5)° . 36.x dx

A: 7) the given integral is (3x4-5)536x3 3) the given integral is (x3-2)-43x2.

Q: dx Vx2 - 1

A:

Q: 42V dx Vx

A: Use the formula, ∫axdx=axlog a+cwhere, c=constant of integrationlog a≠0

Q: evi dx 3D?

A:

Q: 3 т2 dm )2 т3-2

A: NOTE: Refresh your page if you can't see any equations. . ...............(1) substitute…

Q: Evaluate dx

A:

Q: (5a.4) dn 1 (2,2) (a?-x +1

A:

Q: + wha x+8'

A: