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liq [111]
2 years ago
10

A popular chain of superstores made 3.0479×108 dollars in profit last year. One particular store in the chain made 2.102×106 dol

lars in profit last year. How many times greater is the profit made by the entire chain last year than by this particular store? Drag and drop the values into the boxes to represent the answer in scientific notation. 0.145
1.45
145
​10²​​​
10^14​
​10^48​
Mathematics
1 answer:
ollegr [7]2 years ago
7 0
This is a simple exercise with rates.
You just have to divide the total profit of the store, that year, per the profit of the particular store.


[3.0479×10⁸ dollars] ÷ [2.102×10⁶ dollars] = 1,45 x 10²





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Answer:

You brought 3 lbs of  potatoes

Step-by-step explanation:

divide 3.45 by 1.15

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A student bought two juice pouches for $1.25 each and 3 bags of chips. The total cost was $5.05. Write and solve an equation to
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Answer:

$0.85

Step-by-step explanation:

2 juice pouches- $2.50

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1 year ago
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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.

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Alex
The only ones that would work are if the factor is composite and if 1 is added to it it can be divided into 48. So 15 would work. The factor 15 is composite, and if 1 is added it can be divided into 48.Therefore the son is 15.
7 0
2 years ago
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