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mars1129 [50]
4 years ago
9

Jane is taking two courses. The probability she passes the first course is 0.7. The probability she passes the second course is

0.8. The probability she passes at least one of the courses is 0.9. a. What is the probability she passes both courses?
Mathematics
1 answer:
-Dominant- [34]4 years ago
4 0

Answer: p = 0.56

Step-by-step explanation: from the question, probability she passed the first course = 0.7

Probability she passed the second course = 0.8

Probability she passes both courses means probability she passes the first course "and" probability she passes the second course.

Probability she pass both courses = 0.7 * 0.8 = 0.56

Note that "and" in probability means multiplication.

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Step-by-step explanation:

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3 years ago
Factorise the following.​
zepelin [54]

Answer:

Step-by-step explanation:

i). \frac{48a^3}{6a}=\frac{6\times 8\times a\times a^2}{6a}

          =\frac{6a}{6a}\times 8a^2

          = 8a²

ii). \frac{72a^3b^4c^5}{8ab^2c^3} = \frac{8\times 9\times a\times a^2\times b^2\times b^2\times c^2\times c^3}{8ab^2c^3}

                = \frac{8ab^2c^3}{8ab^2c^3}\times 9a^2b^2c^2

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8 0
3 years ago
A restaurant offered cooking classes on 24 of the 30 days in November. What decimal is equivalent to the fraction of days in Nov
zvonat [6]
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3 0
3 years ago
Find the value of x and the length of both chords if BW=2x+10, WD=4, CW=7x+5, and WE=2.
daser333 [38]

Answer:

The value of x is 5

The lengths of the chords are 24 units and 42 units

Step-by-step explanation:

In a circle if two chords intersected at a point inside it there are four segments created, two in each cord, the products of the lengths of the line segments on each chord are equal

∵ BD and CE are two chords in a circle intersected at W

∴ The two segments of chord BD are BW and WD

∴ The two segments of chord CE are CW and WE

- By using the rule above

∴ BW × WD = CW × WE

∵ BW = 2x + 10 and WD = 4

∵ CW = 7x + 5 and WE = 2

- Substitute them in the rule above

∴ (2x + 10) × 4 = (7x + 5) × 2

∴ 4(2x) + 4(10) = 2(7x) + 2(5)

∴ 8x + 40 = 14x + 10

- Subtract 14x from both sides

∴ - 6x + 40 = 10

- Subtract 40 from both sides

∴ - 6x = - 30

- Divide both sides by - 6

∴ x = 5

∵ Chord BD = 2x + 10 + 4

∴ Chord BD = 2x + 14

- Substitute the value of x to find its length

∴ Chord BD = 2(5) + 14 = 10 + 14

∴ Chord BD = 24 units

∵ Chord CE = 7x + 5 + 2

∴ Chord CE = 7x + 7

- Substitute the value of x to find its length

∴ Chord CE = 7(5) + 7 = 35 + 7

∴ Chord CE = 42 units

5 0
3 years ago
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