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DIA [1.3K]
3 years ago
9

Kenyon ran for 8 miles each week while training. Here is his record of the number of miles he ran.

Mathematics
1 answer:
diamong [38]3 years ago
8 0

Step-by-step explanation:

Let x be the miles Kenyon ran on tuesday

Let y be the miles Kenyon ran on thursday

Given,

Mon + Tues + Wed + Thu = 8mi

1.2 + x + 1.6 + y = 8 \\ 2.8 + x + y = 8 \\ x + y = 8 - 2.8 \\ x + y = 5.2

also given,

y = x + 0.4 \\  - x + y = 0.4

Now we have x + y = 5.2 as equation 1 and -x + y = 0.4 as equation 2

Using equation 1,

x + y = 5.2 \\ x = 5.2 - y

Substitute this equation into equation 2.

- (5.2 - y) + y = 0.4 \\  - 5.2 + y + y = 0.4 \\  - 5.2 + 2y = 0.4 \\ 2y = 0.4 + 5.2 \\ 2y = 5.6 \\ y = 5.6 \div 2 \\  = 2.8mi

Substitute y = 2.8 into equation 1.

x + 2.8 = 5.2 \\ x = 5.2 - 2.8 \\  = 2.4mi

Therefore Kenyon ran 2.4 miles on tuesday, and ran 2.8 miles on thursday.

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5^(x+7)=(1/625)^(2x-13)


We move all terms to the left:


5^(x+7)-((1/625)^(2x-13))=0



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Suppose it is known that 60% of radio listeners at a particular college are smokers. A sample of 500 students from the college i
vladimir1956 [14]

Answer:

The probability that at least 280 of these students are smokers is 0.9664.

Step-by-step explanation:

Let the random variable <em>X</em> be defined as the number of students at a particular college who are smokers

The random variable <em>X</em> follows a Binomial distribution with parameters n = 500 and p = 0.60.

But the sample selected is too large and the probability of success is close to 0.50.

So a Normal approximation to binomial can be applied to approximate the distribution of X if the following conditions are satisfied:

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2. n(1 - p) ≥ 10

Check the conditions as follows:

 np=500\times 0.60=300>10\\n(1-p)=500\times(1-0.60)=200>10

Thus, a Normal approximation to binomial can be applied.

So,  

X\sim N(\mu=600, \sigma=\sqrt{120})

Compute the probability that at least 280 of these students are smokers as follows:

Apply continuity correction:

P (X ≥ 280) = P (X > 280 + 0.50)

                   = P (X > 280.50)

                   =P(\frac{X-\mu}{\sigma}>\frac{280-300}{\sqrt{120}}\\=P(Z>-1.83)\\=P(Z

*Use a <em>z</em>-table for the probability.

Thus, the probability that at least 280 of these students are smokers is 0.9664.

8 0
3 years ago
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