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Flura [38]
3 years ago
14

Which of the following is a composite number? A.87 B.31 C.59 D.41

Mathematics
1 answer:
kicyunya [14]3 years ago
8 0
87 is a composite number because it is divisible by 3, all of the rest of the numbers are prime
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tiny-mole [99]

Answer:

30

Step-by-step explanation:

6 0
3 years ago
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Y=(1/2)^x +2. What is the horizontal asymptote of this function?
professor190 [17]
\lim_{x\to -\infty} \left(\frac{1}{2}\right)^x+2=\infty+2=\infty\\
\lim_{x\to \infty} \left(\frac{1}{2}\right)^x+2=0+2=2\\

There is one-sided horizontal asymptote y=2
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3 years ago
Are the triangles similar? If so, state the similarity and the postulate or theorem that justifies your answer.
malfutka [58]

The triangles are similar by SAS principle.

<h3>How to know similar triangles?</h3>

Similar triangles have the same shape but may have different sizes.

In similar triangles, corresponding sides are always in the same ratio.

The corresponding angles are congruent.

Therefore, using SAS ratio,

6 / 8 = 8 × 3 / 32

6 / 8  = 24 / 32 = 3 / 4

Therefore, the corresponding sides are a ratio of each other.

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learn more on similar triangle here: brainly.com/question/21480885

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6 0
2 years ago
This problem has been solved!See the answerA municipal bond service has three rating categories (A, B, and C). Suppose that in t
mariarad [96]

Answer:

a. \frac{35}{51}

b. \frac{51}{100}

c. \frac{1}{5}

Step-by-step explanation:

Suppose cities represented by C', suburbs represented by S and rural represented by R,

Let x be the total number of bonds issued throughout the US,

According to the question,

n(A) = 70% of x = 0.7x,

n(B) = 10% of x = 0.1x,

n(C) = 20% of x = 0.2x,

n(A∩C') = 50% of n(A) = 0.5 × 0.7x = 0.35x,

n(A∩S) = 20% of n(A) = 0.2 × 0.7x = 0.14x,

n(A∩R) = 30% of n(A) = 0.3 × 0.7x = 0.21x,

n(B∩C') = 40% of n(B) = 0.4 × 0.1x = 0.04x,

n(B∩S) = 30% of n(B) = 0.3 × 0.1x = 0.03x,

n(B∩R) = 30% of n(B) = 0.3 × 0.1x = 0.03x,

n(C∩C') = 60% of n(C) = 0.6 × 0.2x = 0.12x,

n(C∩S) = 15% of n(C) = 0.15 × 0.2x = 0.03x,

n(C∩R) = 25% of n(C) = 0.25 × 0.2x = 0.05x,

n(C') = n(A∩C')  + n(B∩C')  + n(C∩C')  = 0.35x + 0.04x + 0.12x = 0.51x

n(S) = n(A∩S) + n(B∩S) + n(C∩S) = 0.14x + 0.03x + 0.03x = 0.20x

a. The probability that it will receive an A rating, if a new municipal bond is to be issued by a city,

P(\frac{A}{C'})=\frac{P(A\cap C')}{P(C')}=\frac{0.35x/x}{0.51x/x}=\frac{0.35}{0.51}=\frac{35}{51}

b. The proportion of municipal bonds are issued by cities = \frac{n(C')}{x}

=\frac{0.51x}{x}

=\frac{51}{100}

c. The proportion of municipal bonds are issued by suburbs = \frac{n(S)}{x}

=\frac{0.20x}{x}

=\frac{20}{100}

=\frac{1}{5}

3 0
3 years ago
(A3- 2a2) - ( 3a2- 4a3)
Tanzania [10]
If you would like to calculate (a^3 - 2 * a^2) - (3 * a^2 - 4 * a^3), you can do this using the following steps:

(a^3 - 2 * a^2) - (3 * a^2 - 4 * a^3) = a^3 - 2 * a^2 - 3 * a^2 + 4 * a^3 = 5 * a^3 - 5 * a^2

The correct result would be 5 * a^3 - 5 * a^2.
8 0
3 years ago
Read 2 more answers
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