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Find the cardinality of the set A=x | -3 < x < 33.

Mathematics

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C

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Part 4· 2 steps

To find the cardinality of the set A={x:−3<x<33}A = \{x : -3 < x < 33\}, we need to count the number of integers between -3 and 33 (exclusive).
The integers in the set are −2,−1,0,1,…,32-2, -1, 0, 1, \dots, 32.

  1. Identify the smallest and largest integers in the set.

    The smallest integer is −2-2.
    The largest integer is 32.

  2. Calculate the cardinality using the formula

    (Largest integer - Smallest integer) + 1.

    Cardinality=32−(−2)+1\text{Cardinality} = 32 - (-2) + 1 Cardinality=32+2+1\text{Cardinality} = 32 + 2 + 1 Cardinality=35\text{Cardinality} = 35

    The correct option is C.
    The final answer is C. 35\boxed{\text{C. } 35}.

Part 5· 3 steps

Given that cos⁡θ=sin⁡70∘\cos \theta = \sin 70^\circ where θ\theta is an acute angle.

  1. Use the trigonometric identity cos⁡θ=sin⁡(90∘−θ)\cos \theta = \sin (90^\circ - \theta).

    So, we can write the given equation as:

    sin⁡(90∘−θ)=sin⁡70∘\sin (90^\circ - \theta) = \sin 70^\circ
  2. Since θ\theta is an acute angle, 90∘−θ90^\circ - \theta must also be an acute angle.

    Therefore, we can equate the angles.

    90∘−θ=70∘90^\circ - \theta = 70^\circ
  3. Solve for θ\theta.

    θ=90∘−70∘\theta = 90^\circ - 70^\circ θ=20∘\theta = 20^\circ

    The correct option is C.
    The final answer is C. 20∘\boxed{\text{C. } 20^\circ}.

Part 6· 3 steps

A fair die is rolled once. We need to find the probability of obtaining a 3 or 4.

  1. Identify the total possible outcomes when rolling a fair die.

    The possible outcomes are {1,2,3,4,5,6}\{1, 2, 3, 4, 5, 6\}.
    Total number of outcomes = 6.

  2. Identify the favorable outcomes (obtaining a 3 or 4).

    The favorable outcomes are {3,4}\{3, 4\}.
    Number of favorable outcomes = 2.

  3. Calculate the probability.

    P(3 or 4)=Number of favorable outcomesTotal number of outcomesP(\text{3 or 4}) = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} P(3 or 4)=26P(\text{3 or 4}) = \frac{2}{6} P(3 or 4)=13P(\text{3 or 4}) = \frac{1}{3}

    The correct option is A.
    The final answer is A. 13\boxed{\text{A. } \frac{1}{3}}.

Part 7· 3 steps

A bag contains two black, three green, and four red balls. A ball is picked at random. We need to find the probability that it is either black or red.

  1. Determine the total number of balls in the bag.

    Number of black balls = 2
    Number of green balls = 3
    Number of red balls = 4

    Total number of balls=2+3+4=9\text{Total number of balls} = 2 + 3 + 4 = 9
  2. Determine the number of favorable outcomes (balls that are either black or red).

    Number of black or red balls=Number of black balls+Number of red balls\text{Number of black or red balls} = \text{Number of black balls} + \text{Number of red balls} Number of black or red balls=2+4=6\text{Number of black or red balls} = 2 + 4 = 6
  3. Calculate the probability.

    P(black or red)=Number of black or red ballsTotal number of ballsP(\text{black or red}) = \frac{\text{Number of black or red balls}}{\text{Total number of balls}} P(black or red)=69P(\text{black or red}) = \frac{6}{9} P(black or red)=23P(\text{black or red}) = \frac{2}{3}

    The correct option is B.
    The final answer is B. 23\boxed{\text{B. } \frac{2}{3}}.

Part 8· 3 steps

Find the sum to infinity of a Geometric Progression whose first term is 10 and common ratio 23\frac{2}{3}.

  1. Recall the formula for the sum to infinity of a Geometric Progression.

    The sum to infinity S∞S_\infty is given by S∞=a1−rS_\infty = \frac{a}{1-r}, where aa is the first term and rr is the common ratio, provided ∣r∣<1|r| < 1.

  2. Identify the given values.

    First term a=10a = 10.
    Common ratio r=23r = \frac{2}{3}.
    Since ∣23∣<1|\frac{2}{3}| < 1, the sum to infinity exists.

  3. Substitute the values into the formula and calculate.

    S∞=101−23S_\infty = \frac{10}{1 - \frac{2}{3}} S∞=1033−23S_\infty = \frac{10}{\frac{3}{3} - \frac{2}{3}} S∞=1013S_\infty = \frac{10}{\frac{1}{3}} S∞=10×3S_\infty = 10 \times 3 S∞=30S_\infty = 30

    The correct option is C.
    The final answer is C. 30\boxed{\text{C. } 30}.

Part 9· 3 steps

The length and breadth of a rectangular piece of land are (x+2)(x+2)m and (x−3)(x-3)m respectively. The perimeter of the land is 30 m. We need to find the value of xx.

  1. Write down the given dimensions and the perimeter formula for a rectangle.

    Length L=(x+2)L = (x+2) m
    Breadth B=(x−3)B = (x-3) m
    Perimeter P=2(L+B)P = 2(L+B)
    Given P=30P = 30 m.

  2. Substitute the expressions for length and breadth into the perimeter formula.

    30=2((x+2)+(x−3))30 = 2((x+2) + (x-3))
  3. Simplify and solve for xx.

    30=2(x+2+x−3)30 = 2(x+2+x-3) 30=2(2x−1)30 = 2(2x-1)

    Divide both sides by 2:

    302=2x−1\frac{30}{2} = 2x-1 15=2x−115 = 2x-1

    Add 1 to both sides:

    15+1=2x15 + 1 = 2x 16=2x16 = 2x

    Divide both sides by 2:

    x=162x = \frac{16}{2} x=8x = 8

    The correct option is A.
    The final answer is A. 8 m\boxed{\text{A. } 8 \text{ m}}.

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