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Irina-Kira [14]
3 years ago
11

Types of bonds are divided into three categories: good risk, medium risk, and poor risk. Assume that of a total of 11, 332 bonds

, 7,311 are good risk, 1,182 are medium risk, and the rest are poor risk. One bond is chosen at random. What is the probability that the bond is not a poor ? Write only a number as your answer. Round to two decimal places (for example : 0.43). Do not write as a percentage .
Mathematics
1 answer:
Alekssandra [29.7K]3 years ago
3 0

Answer:

Probabilty of not poor= 0.75

Step-by-step explanation:

total of 11332 bonds.

7311 are good risk.

1182 are medium risk.

Poor risk

= total risk-(good risk+ medium risk)

= 11332-(7311+1182)

= 11332-8493

= 2839.

Poor risk = 2839

Probabilty that the ball choosen at random is not poor= 1 - probability that the ball is poor

Probability of poor = 2839/11332

Probabilty of poor= 0.2505

Probabilty that the ball choosen at random is not poor= 1- 0.2505

= 0.7495

To two decimal place= 0.75

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None of the choices equal 1/50, so let's just make it a percentage anyway.

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<em>The answer is 2%, or 1/50 as a fraction.</em>

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Choose the correct simplification of (4x − 3)(3x2 − 4x − 3).
tester [92]
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=4x\cdot \:3x^2+4x\left(-4x\right)+4x\left(-3\right)-3\cdot \:3x^2-3\left(-4x\right)-3\left(-3\right) 

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8 0
3 years ago
1 point
swat32

Answer:

\log_{10}(147) = 2.1673

Step-by-step explanation:

Given

\log_{10} 3 = 0.4771

\log_{10} 5 = 0.6990

\log_{10} 7= 0.8451

\log_{10} 11 = 1.0414

Required

Evaluate \log_{10}(147)

Expand

\log_{10}(147) = \log_{10}(49 * 3)

Further expand

\log_{10}(147) = \log_{10}(7 * 7 * 3)

Apply product rule of logarithm

\log_{10}(147) = \log_{10}(7) + \log_{10}(7) + \log_{10}(3)

Substitute values for log(7) and log(3)

\log_{10}(147) = 0.8451 + 0.8451 + 0.4771

\log_{10}(147) = 2.1673

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2 years ago
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