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disa [49]
4 years ago
12

Answer the question based on the data in the two-way table.

Mathematics
2 answers:
Alex777 [14]4 years ago
8 0

Answer:

<h3><u><em>C.</em></u> P(boy|above average grades) P(above average grades)</h3>

Step-by-step explanation:

lisabon 2012 [21]4 years ago
5 0

Answer:

b

Step-by-step explanation:

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4 years ago
Hexagon DEFGHI is translated on the coordinate plane below to create hexagon D'E'F’G’H’I’
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The answer is "\bold{(x,y)\longrightarrow (x-8, y-7)}".

Step-by-step explanation:

In the given question some data is missing. so, the correct answer to this can be explained in the following example:

In the given example, point D  is also known as the coordinates in the (2, 5), and in the coordination of the D prime (D') and after translation its co-ordinates value is (-6, -2). Since the very first choice is only accurate as:

⇒2-8 = -6

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7 0
3 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
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2 years ago
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