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Nitella [24]
3 years ago
6

Two distinct integers, $x$ and $y$, are randomly chosen from the set $\{1,2,3,4,5,6,7,8,9,10\}$. What is the probability that $x

y-x-y$ is even?
Mathematics
1 answer:
Vaselesa [24]3 years ago
7 0
24 i really dont know what 

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lamars bowling scores were 124, 150, and 161, during league competition. if lamars previous bowling average was a 129, how much
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What are the values of a and b
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7 0
3 years ago
A quality-conscious disk manufacturer wishes to know the fraction of disks his company makes which are defective. Step 2 of 2 :
const2013 [10]

Answer:

The 80% confidence interval for the population proportion of disks which are defective is (0.059, 0.079).

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of \pi, and a confidence level of 1-\alpha, we have the following confidence interval of proportions.

\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}

In which

z is the zscore that has a pvalue of 1 - \frac{\alpha}{2}.

Suppose a sample of 1067 floppy disks is drawn. Of these disks, 74 were defective.

This means that n = 1067, \pi = \frac{74}{1067} = 0.069

80% confidence level

So \alpha = 0.2, z is the value of Z that has a pvalue of 1 - \frac{0.2}{2} = 0.9, so Z = 1.28.

The lower limit of this interval is:

\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.069 - 1.28\sqrt{\frac{0.069*0.931}{1067}} = 0.059

The upper limit of this interval is:

\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.069 + 1.28\sqrt{\frac{0.069*0.931}{1067}} = 0.079

The 80% confidence interval for the population proportion of disks which are defective is (0.059, 0.079).

7 0
2 years ago
What is a quotient of 3/10÷8/5 out of these forms 1. 3/16 2. 4/9 3. 15/80 4. 24/50
Zanzabum

Answer:

option 1

Step-by-step explanation:

\frac{3}{10} ÷ \frac{8}{5}

• leave the first fraction

• change division to multiplication

•  turn the second fraction 'upside down'

= \frac{3}{10} × \frac{5}{8} ( cancel 5 and 10 by 5 )

= \frac{3}{2} × \frac{1}{8}

= \frac{3}{16}

8 0
2 years ago
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