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pogonyaev
3 years ago
12

Which of the following is an example of a rational number?

Mathematics
1 answer:
ivolga24 [154]3 years ago
6 0
<h3>Answer: Choice B</h3>

==================================================

Explanation:

Using your calculator, you should find that

  • sqrt(125) = 11.1803 approximately
  • sqrt(196) = 14 exactly
  • sqrt(207) = 14.387 approximately
  • sqrt(220) = 14.832 approximately

Each approximate result has its decimal digits go on forever without any pattern. So we cannot write those values as a fraction of two integers. The decimal expansion must have some kind of pattern that repeats forever, so we could consider it rational.

Only choice B results in a whole number. The value 14 is the same as 14/1, showing it can be written as a fraction of two integers. Therefore choice B is rational.

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Complete each statement. An event with a probability of 0 is An event with a probability of 1 is
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Answer:

An event with a probability of 0 is impossible.  

An event with a probability of 1 is certain.

Step-by-step explanation:

Probability is typically expressed in terms of a fraction between 0 and 1 where the denominator is the total number of outcomes and the numerator is the number of desired outcomes.  Since probability is expressed as a fraction, if the probability is 0, that means it is impossible, or there is no chance that the event can happen.  However, if the probability is 1, that means that the event is certain to happen and the odds are completely in your favor that the event will happen.  

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3 years ago
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The range of a data set is 18. if the smallest number in the set is 9, What is the largest number?
tankabanditka [31]
Range is the difference between the largest and the smallest variable. If the range is 18 and the smallest number is 9. Add 9 to 18.
9+18= 27. 27 is the largest number.
7 0
3 years ago
In a process that manufactures bearings, 90% of the bearings meet a thickness specification. A shipment contains 500 bearings. A
Marina86 [1]

Answer:

(a) 0.94

(b) 0.20

(c) 90.53%

Step-by-step explanation:

From a population (Bernoulli population), 90% of the bearings meet a thickness specification, let p_1 be the probability that a bearing meets the specification.

So, p_1=0.9

Sample size, n_1=500, is large.

Let X represent the number of acceptable bearing.

Convert this to a normal distribution,

Mean: \mu_1=n_1p_1=500\times0.9=450

Variance: \sigma_1^2=n_1p_1(1-p_1)=500\times0.9\times0.1=45

\Rightarrow \sigma_1 =\sqrt{45}=6.71

(a) A shipment is acceptable if at least 440 of the 500 bearings meet the specification.

So, X\geq 440.

Here, 440 is included, so, by using the continuity correction, take x=439.5 to compute z score for the normal distribution.

z=\frac{x-\mu}{\sigma}=\frac{339.5-450}{6.71}=-1.56.

So, the probability that a given shipment is acceptable is

P(z\geq-1.56)=\int_{-1.56}^{\infty}\frac{1}{\sqrt{2\pi}}e^{\frac{-z^2}{2}}=0.94062

Hence,  the probability that a given shipment is acceptable is 0.94.

(b) We have the probability of acceptability of one shipment 0.94, which is same for each shipment, so here the number of shipments is a Binomial population.

Denote the probability od acceptance of a shipment by p_2.

p_2=0.94

The total number of shipment, i.e sample size, n_2= 300

Here, the sample size is sufficiently large to approximate it as a normal distribution, for which mean, \mu_2, and variance, \sigma_2^2.

Mean: \mu_2=n_2p_2=300\times0.94=282

Variance: \sigma_2^2=n_2p_2(1-p_2)=300\times0.94(1-0.94)=16.92

\Rightarrow \sigma_2=\sqrt(16.92}=4.11.

In this case, X>285, so, by using the continuity correction, take x=285.5 to compute z score for the normal distribution.

z=\frac{x-\mu}{\sigma}=\frac{285.5-282}{4.11}=0.85.

So, the probability that a given shipment is acceptable is

P(z\geq0.85)=\int_{0.85}^{\infty}\frac{1}{\sqrt{2\pi}}e^{\frac{-z^2}{2}=0.1977

Hence,  the probability that a given shipment is acceptable is 0.20.

(c) For the acceptance of 99% shipment of in the total shipment of 300 (sample size).

The area right to the z-score=0.99

and the area left to the z-score is 1-0.99=0.001.

For this value, the value of z-score is -3.09 (from the z-score table)

Let, \alpha be the required probability of acceptance of one shipment.

So,

-3.09=\frac{285.5-300\alpha}{\sqrt{300 \alpha(1-\alpha)}}

On solving

\alpha= 0.977896

Again, the probability of acceptance of one shipment, \alpha, depends on the probability of meeting the thickness specification of one bearing.

For this case,

The area right to the z-score=0.97790

and the area left to the z-score is 1-0.97790=0.0221.

The value of z-score is -2.01 (from the z-score table)

Let p be the probability that one bearing meets the specification. So

-2.01=\frac{439.5-500  p}{\sqrt{500 p(1-p)}}

On solving

p=0.9053

Hence, 90.53% of the bearings meet a thickness specification so that 99% of the shipments are acceptable.

8 0
3 years ago
W/5=3 what does w equal
m_a_m_a [10]

Answer:

w=15

Step-by-step explanation:

w/5=3

multiply by 5 on both sides

w=3*5

∴ w=15

7 0
3 years ago
Read 2 more answers
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