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FrozenT [24]
3 years ago
11

Solve for p: 7p + 15.89 = 19.32

Mathematics
1 answer:
Bond [772]3 years ago
4 0

Answer:

p=0.07

Step-by-step explanation:

7p+15.89=19.32

19.32-15.89=3.43

3.43/7= 0.07

p=0.07

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50% markup of $111.00 is the same as increasing 111 by half

50% of 111 = 55.5
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Final value $166.50
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Type the ordered pair for each of the lettered points on the graph. Use parentheses () when typing the answer.
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A=(1,5) E=(-1,-4) I=(-2,4)
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Mark the points with the coordinates (4, 14), (22, 6), and (16, 18). Connect the points to form a triangle. and what type of tri
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Answer:

Step-by-step explanation:

You have  the  points (4, 14), (22, 6), and (16, 18).

It is important to remember that he first number of each point is the x-coordinate of that point and the second number of each one of them is the y-coordinate of that point.

Therefore, knowing the above, you can mark each point, as you can observe in the image attached, and then you can connect the points to form the triangle shown in the image.

3 0
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Read 2 more answers
PLEASE HELP (80 points!!) <br> What if the following are equivalent to x^2/5 (more than once answer)
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Answer:

\sqrt[5]{x^2} and (\sqrt[5]{x} )^2

Step-by-step explanation:

They are both correct.

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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
4 years ago
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