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Which of the following formulas best describes the formula for standard of a grouped data?

Mathematics

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Answer

C

Step-by-step solution

31 steps

Here are the solutions to the questions:

Part 9· 3 steps

To find the probability of getting at least one head in a toss of two coins, list all possible outcomes.

  1. List all possible outcomes when tossing two coins.

    The possible outcomes are: (Head, Head), (Head, Tail), (Tail, Head), (Tail, Tail).
    Total number of outcomes = 4.

  2. Identify the outcomes with at least one head.

    The outcomes with at least one head are: (Head, Head), (Head, Tail), (Tail, Head).
    Number of favorable outcomes = 3.

  3. Calculate the probability.

    Probability=Number of favorable outcomesTotal number of outcomes=34\text{Probability} = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{3}{4}

    The correct option is C.

    C) 34\boxed{\text{C) } \frac{3}{4}}

Part 10· 3 steps

Find the missing term in the pattern 2, 3, 7, 10, 15, ..., 32.
Let's analyze the differences between consecutive terms:
3−2=13 - 2 = 1
7−3=47 - 3 = 4
10−7=310 - 7 = 3
15−10=515 - 10 = 5
The sequence of differences is 1, 4, 3, 5.
This pattern suggests an alternating sequence of differences:
The odd-positioned differences are 1,3,5,…1, 3, 5, \dots (increasing by 2).
The even-positioned differences are 4,…4, \dots
If we assume the next odd-positioned difference is 7 (after 5), and the next even-positioned difference is 10 (after 4, assuming 4,104, 10 is part of a sequence 4,7,104, 7, 10 or 4,6,8,104, 6, 8, 10 or 4,5,6,7,8,9,104, 5, 6, 7, 8, 9, 10).
Let's try the pattern of differences: 1,4,3,5,7,101, 4, 3, 5, 7, 10.

  1. Apply the differences to find the terms.

    a1=2a_1 = 2
    a2=2+1=3a_2 = 2 + 1 = 3
    a3=3+4=7a_3 = 3 + 4 = 7
    a4=7+3=10a_4 = 7 + 3 = 10
    a5=10+5=15a_5 = 10 + 5 = 15

  2. Find the missing term (a6a_6) using the next difference in the pattern (which would be 7).

    a6=15+7=22a_6 = 15 + 7 = 22

  3. Check if the next term (a7a_7) matches the given last term (32) using the next difference (which would be 10).

    a7=22+10=32a_7 = 22 + 10 = 32
    This pattern perfectly matches the given sequence and the last term.
    The missing term is 22.
    The correct option is B.

    B) 22\boxed{\text{B) } 22}

Part 11· 4 steps

Find the percentile rank of 6 for the data set: 2, 3, 5, 6, 7, 7, 9, 12.

  1. Arrange the data in ascending order (it is already arranged).

    Data set: 2, 3, 5, 6, 7, 7, 9, 12.

  2. Count the total number of data points (NN).

    N=8N = 8.

  3. Count the number of data points less than or equal to 6 (CC).

    The data points less than or equal to 6 are 2, 3, 5, 6.
    C=4C = 4.

  4. Calculate the percentile rank using the formula

    Percentile Rank=Number of values less than or equal to xTotal number of values×100\text{Percentile Rank} = \frac{\text{Number of values less than or equal to } x}{\text{Total number of values}} \times 100.

    Percentile Rank=48×100%=12×100%=50%\text{Percentile Rank} = \frac{4}{8} \times 100\% = \frac{1}{2} \times 100\% = 50\%

    The correct option is C.

    C) 50%\boxed{\text{C) } 50\%}

Part 12· 2 steps

Convert 2π3\frac{2\pi}{3} radians to degrees.

  1. Use the conversion factor 1 radian=180∘π1 \text{ radian} = \frac{180^\circ}{\pi}.

    2π3 radians=2π3×180∘π\frac{2\pi}{3} \text{ radians} = \frac{2\pi}{3} \times \frac{180^\circ}{\pi}

  2. Simplify the expression.

    =2×180∘3= \frac{2 \times 180^\circ}{3} =2×60∘= 2 \times 60^\circ =120∘= 120^\circ

    The correct option is B.

    B) 120∘\boxed{\text{B) } 120^\circ}

Part 13· 2 steps

Evaluate the definite integral ∫13(3x2+2x+5) dx\int_{1}^{3} (3x^2 + 2x + 5)\,dx.

  1. Find the antiderivative of the function.

    ∫(3x2+2x+5) dx=3x33+2x22+5x+C=x3+x2+5x+C\int (3x^2 + 2x + 5)\,dx = 3\frac{x^3}{3} + 2\frac{x^2}{2} + 5x + C = x^3 + x^2 + 5x + C

  2. Evaluate the antiderivative at the upper and lower limits and subtract.

    [x3+x2+5x]13=(33+32+5×3)−(13+12+5×1)[x^3 + x^2 + 5x]_{1}^{3} = (3^3 + 3^2 + 5 \times 3) - (1^3 + 1^2 + 5 \times 1) =(27+9+15)−(1+1+5)= (27 + 9 + 15) - (1 + 1 + 5) =51−7= 51 - 7 =44= 44

    The correct option is B.

    B) 44\boxed{\text{B) } 44}

Part 14· 4 steps

A student has gotten the following grades on his test: 87, 95, 76, and 88. He wants an 85 or better overall. Find the minimum grade he must get on the last test to achieve that average.

  1. Let xx be the grade on the last test.

    The total number of tests is 5.

  2. Set up the inequality for the average grade.

    87+95+76+88+x5≥85\frac{87 + 95 + 76 + 88 + x}{5} \ge 85

  3. Sum the known grades.

    87+95+76+88=34687 + 95 + 76 + 88 = 346

  4. Substitute the sum into the inequality and solve for xx.

    346+x5≥85\frac{346 + x}{5} \ge 85 346+x≥85×5346 + x \ge 85 \times 5 346+x≥425346 + x \ge 425 x≥425−346x \ge 425 - 346 x≥79x \ge 79

    The minimum grade he must get on the last test is 79.
    The correct option is D.

    D) 79\boxed{\text{D) } 79}

Part 15· 2 steps

Find the exact value of (7−2i)(7+2i)(7-2i)(7+2i).

  1. Recognize this as a product of complex conjugates, which follows the form (a−bi)(a+bi)=a2+b2(a-bi)(a+bi) = a^2 + b^2.

    Here, a=7a=7 and b=2b=2.

  2. Apply the formula.

    (7−2i)(7+2i)=72+22(7-2i)(7+2i) = 7^2 + 2^2 =49+4= 49 + 4 =53= 53

    The correct option is A.

    A) 53\boxed{\text{A) } 53}

Part 16· 2 steps

Find the argument of the complex number 5+2i5 + 2i.

  1. The argument θ\theta of a complex number z=x+yiz = x + yi is given by tan⁡θ=yx\tan \theta = \frac{y}{x}.

    Here, x=5x=5 and y=2y=2.

    tan⁡θ=25\tan \theta = \frac{2}{5}

  2. Calculate θ\theta.

    θ=arctan⁡(25)\theta = \arctan\left(\frac{2}{5}\right)

    The question asks for the exact value, and the options are numerical. Let's check the options.
    A) 5, B) 29\sqrt{29}, C) 31\sqrt{31}, D) 2.
    None of the options represent an angle or arctan⁡(2/5)\arctan(2/5).
    The modulus of the complex number is ∣z∣=x2+y2=52+22=25+4=29|z| = \sqrt{x^2+y^2} = \sqrt{5^2+2^2} = \sqrt{25+4} = \sqrt{29}.
    It seems the question might be asking for the modulus, not the argument, given the options. If it is asking for the modulus, then option B is correct.
    Assuming the question meant to ask for the modulus:

    Modulus ∣z∣=52+22=25+4=29\text{Modulus } |z| = \sqrt{5^2 + 2^2} = \sqrt{25 + 4} = \sqrt{29}

    The correct option, if asking for the modulus, is B.

    B) 29\boxed{\text{B) } \sqrt{29}}

Part 17· 3 steps

Given f(x)=3x3−8x2−5f(x) = 3x^3 - 8x^2 - 5, find f′′(1)f''(1).

  1. Find the first derivative, f′(x)f'(x).

    f′(x)=ddx(3x3−8x2−5)=9x2−16xf'(x) = \frac{d}{dx}(3x^3 - 8x^2 - 5) = 9x^2 - 16x

  2. Find the second derivative, f′′(x)f''(x).

    f′′(x)=ddx(9x2−16x)=18x−16f''(x) = \frac{d}{dx}(9x^2 - 16x) = 18x - 16

  3. Evaluate f′′(x)f''(x) at x=1x=1.

    f′′(1)=18(1)−16=18−16=2f''(1) = 18(1) - 16 = 18 - 16 = 2

    The correct option is C.

    C) 2\boxed{\text{C) } 2}

Part 18

Which of the following formulas best describes the formula for standard deviation of a grouped data?
The standard deviation for grouped data is typically given by the formula:

s=∑f(x−xˉ)2∑fs = \sqrt{\frac{\sum f(x - \bar{x})^2}{\sum f}}

where ff is the frequency, xx is the midpoint of the class interval, and xˉ\bar{x} is the mean.
Let's examine the given options:
A) ∑f(x−xˉ)2n\sqrt{\frac{\sum f(x - \bar{x})^2}{n}} (This is for sample standard deviation, where n=∑fn = \sum f)
B) ∑f(x−xˉ)n\frac{\sum f(x - \bar{x})}{n} (This is not standard deviation)
C) ∑f(x−xˉ)2∑f\sqrt{\frac{\sum f(x - \bar{x})^2}{\sum f}} (This is the population standard deviation for grouped data, or sample standard deviation if ∑f\sum f is n−1n-1)
D) ∑f(x−xˉ)2∑f−1\sqrt{\frac{\sum f(x - \bar{x})^2}{\sum f - 1}} (This is the sample standard deviation for grouped data)

Option C is the formula for the population standard deviation of grouped data. Option D is for the sample standard deviation of grouped data. Often, "standard deviation" without specifying population or sample refers to the sample standard deviation, but option C is also a valid form. Given the options, C is a common representation for standard deviation of grouped data. If nn represents ∑f\sum f, then A and C are similar. However, C explicitly uses ∑f\sum f in the denominator, which is standard for population standard deviation. If it's sample standard deviation, the denominator should be ∑f−1\sum f - 1.
Assuming the question implies population standard deviation or a general formula where n=∑fn = \sum f.
The formula for standard deviation of grouped data is:

s=∑fi(xi−xˉ)2Ns = \sqrt{\frac{\sum f_i (x_i - \bar{x})^2}{N}}

where N=∑fiN = \sum f_i.
Option C matches this form.

C) ∑f(x−xˉ)2∑f\boxed{\text{C) } \sqrt{\frac{\sum f(x - \bar{x})^2}{\sum f}}}

Part 19· 2 steps

Express 14×13×12×1114 \times 13 \times 12 \times 11 as factorial form.

  1. Recall the definition of factorial

    n!=n×(n−1)×⋯×1n! = n \times (n-1) \times \dots \times 1.

  2. To express the given product in factorial form, we need to multiply and divide by the missing terms to complete the factorial.

    14×13×12×11=14×13×12×11×10×9×⋯×110×9×⋯×114 \times 13 \times 12 \times 11 = \frac{14 \times 13 \times 12 \times 11 \times 10 \times 9 \times \dots \times 1}{10 \times 9 \times \dots \times 1} =14!10!= \frac{14!}{10!}

    The correct option is D.

    D) 14!10!\boxed{\text{D) } \frac{14!}{10!}}

Part 20· 4 steps

Simplify 1−cos⁡2x\sqrt{1 - \cos^2 x}.

  1. Use the Pythagorean trigonometric identity sin⁡2x+cos⁡2x=1\sin^2 x + \cos^2 x = 1.

  2. Rearrange the identity to solve for sin⁡2x\sin^2 x.

    sin⁡2x=1−cos⁡2x\sin^2 x = 1 - \cos^2 x

  3. Substitute this into the given expression.

    1−cos⁡2x=sin⁡2x\sqrt{1 - \cos^2 x} = \sqrt{\sin^2 x}

  4. Simplify the square root.

    sin⁡2x=∣sin⁡x∣\sqrt{\sin^2 x} = |\sin x|

    However, typically in multiple-choice questions like this, it's assumed that sin⁡x≥0\sin x \ge 0 or the principal value is taken, so the absolute value is often omitted in the options.
    The options are A) sin⁡x\sin x, B) 1+cos⁡2x\sqrt{1 + \cos 2x}, C) 1, D) sin⁡2x\sin^2 x.
    Given the options, sin⁡x\sin x is the intended answer.
    The correct option is A.

    A) sin⁡x\boxed{\text{A) } \sin x}
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