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Elena-2011 [213]
2 years ago
10

A basketball stadium holds 18500 seats. 65% of the seats are in the lower bowl and 35% of the seats are in the upper bowl. A low

er bowl ticket cost $120 and an upper bowl ticket cost $55. Let x represent the number of lower bowl tickets and y represents the number of upper bowl tickets. What are the restrictions on x and y?
Mathematics
1 answer:
Doss [256]2 years ago
8 0
The restrictions are:


1) x ≥ 0


2) number of lower bowl seats = 65% * 18500 = 12,025

=> x ≤ 12,025
 

3) y ≥ 0


4) number of upper bowl seats = 35% * 18500 = 6,475

=> y ≤ 6,475


It is not necessary to state that x + y ≤ 18,500 because it is included in the other restrictions.
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The average life of a bread-making machine is 7 years, with a standard deviation of 1 year. Assuming that the lives of these mac
Alina [70]

Answer:

a) P(6.4

b) a=7 +1.036*0.333=7.345

So the value of bread-making machine that separates the bottom 85% of data from the top 15% is 7.345.

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  

The central limit theorem states that "if we have a population with mean μ and standard deviation σ and take sufficiently large random samples from the population with replacement, then the distribution of the sample means will be approximately normally distributed. This will hold true regardless of whether the source population is normal or skewed, provided the sample size is sufficiently large".

Let X the random variable life of a bread making machine. We know from the problem that the distribution for the random variable X is given by:

X\sim N(\mu =7,\sigma =1)

We take a sample of n=9 . That represent the sample size.

From the central limit theorem we know that the distribution for the sample mean \bar X is also normal and is given by:

\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})

\bar X \sim N(\mu=7, \frac{1}{\sqrt{9}})

Solution to the problem

Part a

(a) the probability that the mean life of a random sample  of 9 such machines falls between 6.4 and 7.2

In order to answer this question we can use the z score in order to find the probabilities, the formula given by:

z=\frac{\bar X- \mu}{\frac{\sigma}{\sqrt{n}}}

The standard error is given by this formula:

Se=\frac{\sigma}{\sqrt{n}}=\frac{1}{\sqrt{9}}=0.333

We want this probability:

P(6.4

Part b

b) The value of x to the right of which 15% of the  means computed from random samples of size 9 would fall.

For this part we want to find a value a, such that we satisfy this condition:

P(\bar X>a)=0.15   (a)

P(\bar X   (b)

Both conditions are equivalent on this case. We can use the z score again in order to find the value a.  

As we can see on the figure attached the z value that satisfy the condition with 0.85 of the area on the left and 0.15 of the area on the right it's z=1.036. On this case P(Z<1.036)=0.85 and P(Z>1.036)=0.15

If we use condition (b) from previous we have this:

P(X  

P(z

But we know which value of z satisfy the previous equation so then we can do this:

z=1.036

And if we solve for a we got

a=7 +1.036*0.333=7.345

So the value of bread-making machine that separates the bottom 85% of data from the top 15% is 7.345.

8 0
2 years ago
Compare the decimals by using the symbol &lt; or &gt;. a. 0.531 _____ 0.456 b. 0.63 _____ 0.47 c. 0.267 _____ 0.539 d. 0.07 ____
rodikova [14]

Answer:

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0.63 > 0.47

0.267  < 0.539

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Step-by-step explanation:

a. 0.531 _____ 0.456

compare the numbers from left to right

5 is greater than 4 so 0.531 greater than 0.456

0.531  > 0.456

b. 0.63 _____ 0.47

6 is greater than 4. 0.63 > 0.47

c. 0.267 _____ 0.539

2 is less than 5 . so 0.267 less than 0.539

0.267  < 0.539

d. 0.07 _____ 0.7

0 is less than 7

0.07 < 0.7

6 0
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Step-by-step explanation:

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