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svlad2 [7]
3 years ago
14

In a batch of 280 water purifiers, 12 were found to be defective. What is the probability that a water purifier chosen at random

will be defective? Write the probability as a percent. Round to the nearest tenth of a percent if necessary
Mathematics
2 answers:
aleksley [76]3 years ago
6 0
This is experimental probability.

If 12 were found to be defective out of 280 the experimental probability of a purifier being defective is:

12/280 which is:

4.3%  (to nearest tenth of a percent)
Gnom [1K]3 years ago
3 0

Answer:

 \text{Probability}=4.3\%

Step-by-step explanation:

Given : In a batch of 280 water purifiers, 12 were found to be defective.

To find : What is the probability that a water purifier chosen at random will be defective?  Write the probability as a percent.

Solution :

Total number of batch of purifiers = 280

Number of defective purifiers = 12

The probability that a water purifier chosen at random will be defective is given by,

\text{Probability}=\frac{\text{Favorable outcome}}{\text{Total outcome}}

 \text{Probability}=\frac{12}{280}

 \text{Probability}=\frac{3}{70}

Converting into percentage,

 \text{Probability}=\frac{3}{70}\times 100

 \text{Probability}=4.28\%

Round to nearest tenths,

 \text{Probability}=4.3\%

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<h3>What is the meaning of "the power of ten"?</h3>

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<h3>What region has a population with seven as the power of ten?</h3>

From the options, the only region that has as a exponent the number seven is Oceania.

Note: This question is incomplete; here is the missing information:

  • In the introduction, we looked at the human populations of different geographic groups. There are about 1 x 10^9 people in Africa, 4.2 x 10^9 in Asia, 7.4 x 10^8 in Northern America, 3.7 x 10^7 in Oceania, and 6 x 10^8 in Latin America and the Caribbean.

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Please Solve this, it would be extremely helpful for me.
Sidana [21]

Answer:

Step-by-step explanation:

1) ΔCPD & ΔEPF

∠CPD = ∠EPF   { Vertically opposite angles}

∠CDP = ∠PFE {CD║EF, FD is transversal, Alternate interior angles are equal}

ΔCPD ≈ΔEPF  {AA criteria for similarity }

\frac{DC}{EF} =\frac{PC}{EP}\\\\\\\frac{27}{EF}=\frac{15}{7.5}\\\\

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EF =\frac{27*7.5}{15}\\\\

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ii) EF // AB, so Triangles ACB & ECF are similar triangles

\frac{AB}{EF}=\frac{AC}{EC}\\\\\frac{22.5}{13.5}=\frac{AC}{22.5}

AC= \frac{22.5*22.5}{13.5}\\\\AC=37.5 cm

AC = 37.5 cm

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