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madreJ [45]
4 years ago
5

Ryan opens a bank account with $50. Each month he deposits an amount that doubles his savings. Ryan gets $2 from the bank each m

onth as an incentive for maintaining this savings pattern. Which recursive function represents the amount in Ryan’s account in any month?
next = 2 × now + 2, starting at 50
next = 2 + 2, starting at 2
next = now + 2, starting at 50
next = 2 – now, starting at 50
next = 2 × now, starting at 2
Mathematics
2 answers:
tigry1 [53]4 years ago
6 0
<span>next = 2 × now + 2, starting at 50</span>
SOVA2 [1]4 years ago
3 0

Answer:

A.) next = 2 × now + 2, starting at 50

Step-by-step explanation:

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The value of the expression 23+32-3×4-52÷5+(7×4) is
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4 years ago
During the first year, ABC's stock price starts at $ \$100 $ and increases $ 100\% $. During the second year, its stock price go
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Answer:

Step-by-step explanation:

During the first year, ABC's stock price starts at $100 and increases by 100%. This means that the amount by which the stock increased would be

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6 0
4 years ago
Read 2 more answers
The amount of warpage in a type of wafer used in the manufacture of integrated circuits has mean 1.3 mm and standard deviation 0
valina [46]

Answer:

a) P(\bar X >1.305)=P(Z>\frac{1.305-1.3}{\frac{0.1}{\sqrt{200}}}=0.707)

And using the complement rule, a calculator, excel or the normal standard table we have that:

P(Z>0.707)=1-P(Z

b) z=-0.674

And if we solve for a we got

a=1.3 -0.674* \frac{0.1}{\sqrt{200}}=

So the value of height that separates the bottom 95% of data from the top 5% is 1.295.

c) P( \bar X >1.305) = 0.05  

We can use the z score formula:

P( \bar X >1.305) = 1-P(\bar X

Then we have this:

P(z< \frac{1.305-1.3}{\frac{0.1}{\sqrt{n}}}) = 0.95

And a value that accumulates 0.95 of the area on the normal distribution z = 1.64 and we can solve for n like this:

n = (1.64*\frac{0.1}{1.305-1.3})^2= 1075.84 \approx 1076

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 Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations 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".

Part a

Let X the random variable that represent the amount of warpage of a population and we know

Where \mu=1.3 and \sigma=0.1

Since the sample size is large enough we can use the central limit theorem andwe know that the distribution for the sample mean \bar X is given by:

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

We can find the probability required like this:

P(\bar X >1.305)=P(Z>\frac{1.305-1.3}{\frac{0.1}{\sqrt{200}}}=0.707)

And using the complement rule, a calculator, excel or the normal standard table we have that:

P(Z>0.707)=1-P(Z

Part b

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

P(\bar X>a)=0.75   (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 0.25 of the area on the left and 0.75 of the area on the right it's z=-0.674. On this case P(Z<-0.674)=0.25 and P(z>-0.674)=0.75

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=-0.674

And if we solve for a we got

a=1.3 -0.674* \frac{0.1}{\sqrt{200}}=

So the value of height that separates the bottom 95% of data from the top 5% is 1.295.

Part c

For this case we want this condition:

P( \bar X >1.305) = 0.05  

We can use the z score formula:

P( \bar X >1.305) = 1-P(\bar X

Then we have this:

P(z< \frac{1.305-1.3}{\frac{0.1}{\sqrt{n}}}) = 0.95

And a value that accumulates 0.95 of the area on the normal distribution z = 1.64 and we can solve for n like this:

n = (1.64*\frac{0.1}{1.305-1.3})^2= 1075.84 \approx 1076

6 0
3 years ago
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