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statuscvo [17]
3 years ago
10

Malory isn’t putting money in two savings accounts.Account A started with $150 and Account B started with $300.Malory deposits $

16 in Account A and $12 in Account B each month.In how many months will Account A have a balance at least as great as Account B?What will that balance be?
Mathematics
1 answer:
EastWind [94]3 years ago
8 0

Given:

  • There are two savings accounts - account A has $150 and account B has $300 to begin with.
  • Malorie deposits $16 in account A and $12 in account B every month.

To Find:

The number of months where the balance of both is equal and the balance itself.

Solution:

In 37.5 months, the account balances in both accounts will be equal.

This balance will be equal to $750 in both account A and B.

Calculation:

Let x be the number of months where both the accounts have equal balance.

So, at the end of x months, 16x is the amount of money that will be added to account A (since $16 is added every month).

And the total amount at the end of x months will therefore be $(150+16x)

Similarly, at the end of x months, 12x is the amount of money that will be added to account B (since $12 will be added every month).

And the total amount at the end of x months will therefore be $(300+12x)

By our assumption, x is the number of months it will take for both the accont balances will be equal. Therefore, we have the equation

150+16x=300+12x\\4x=150\\x=37.5

Therefore, in 37.5 months, the account balances in both accounts will be equal.

This balance will be equal to 150+16(37.5)=750 i.e. $750 will be the balance in both account A and B.

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The manufacturer of an airport baggage scanning machine claims it can handle an average of530 bags per hour.(a-1) At α = .05 in
-Dominant- [34]

Answer:

b. H0: μ < 530. Reject H0 if tcalc > -1.753

Step-by-step explanation:

1) Data given and notation  

\bar X=507 represent the sample mean

s=47 represent the sample standard deviation  

n=16 sample size  

\mu_o =530 represent the value that we want to test  

\alpha represent the significance level for the hypothesis test.  

t would represent the statistic (variable of interest)  

p_v represent the p value for the test (variable of interest)  

State the null and alternative hypotheses.  

We need to conduct a hypothesis in order to check if the mean is less than 530 (left tailed tes), the system of hypothesis would be:  

Null hypothesis:\mu \geq 530  

Alternative hypothesis:\mu < 530  

If we analyze the size for the sample is < 30 and we don't know the population deviation so is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:  

t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}} (1)  

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".  

Calculate the statistic  

We can replace in formula (1) the info given like this:  

t=\frac{507-530}{\frac{47}{\sqrt{16}}}=-1.957

Critical value

Since we are conducting a left tailed test we need to first find the degrees of freedom for the statistic given by:

df=n-1=16-1=15

Now we need to look on the t distribution with 15 degrees of freedom that accumulates 0.05 of the area on the left area. And on this case the critical value would be t_{\alpha}=-1/753.

And we can use the following excel code to find it: "=T.INV(0.05,15)"  

So then the correct rejection zone for H0 would be: Reject H0 if t_{calc}

P-value  

Since is a one side left tailed test the p value would be:  

p_v =P(t_{(15)}  

Conclusion  

If we compare the p value and the significance level for example \alpha=0.05 we see that p_v so we can conclude that we have enough evidence to fail reject the null hypothesis, so we can conclude that the mean is significantly lower than 530 at 5% of significance.  

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3 years ago
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jenyasd209 [6]
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3 years ago
Find the absolute extrema of the function over the region R. (In each case, R contains the boundaries.) Use a computer algebra s
algol [13]

<em>f(x, y)</em> = <em>x</em> ² - 4<em>xy</em> + 5

has critical points where both partial derivatives vanish:

∂<em>f</em>/∂<em>x</em> = 2<em>x</em> - 4<em>y</em> = 0   ==>   <em>x</em> = 2<em>y</em>

∂<em>f</em>/∂<em>y</em> = -4<em>x</em> = 0   ==>   <em>x</em> = 0   ==>   <em>y</em> = 0

The origin does not lie in the region <em>R</em>, so we can ignore this point.

Now check the boundaries:

• <em>x</em> = 1   ==>   <em>f</em> (1, <em>y</em>) = 6 - 4<em>y</em>

Then

max{<em>f</em> (1, <em>y</em>) | 0 ≤ <em>y</em> ≤ 2} = 6 when <em>y</em> = 0

max{<em>f</em> (1, <em>y</em>) | 0 ≤ <em>y</em> ≤ 2} = -2 when <em>y</em> = 2

• <em>x</em> = 4   ==>   <em>f</em> (4, <em>y</em>) = 12 - 16<em>y</em>

Then

max{<em>f</em> (4, <em>y</em>) | 0 ≤ <em>y</em> ≤ 2} = 12 when <em>y</em> = 0

max{<em>f</em> (4, <em>y</em>) | 0 ≤ <em>y</em> ≤ 2} = -4 when <em>y</em> = 2

• <em>y</em> = 0   ==>   <em>f</em> (<em>x</em>, 0) = <em>x</em> ² + 5

Then

max{<em>f</em> (<em>x</em>, 0) | 1 ≤ <em>x</em> ≤ 4} = 21 when <em>x</em> = 4

min{<em>f</em> (<em>x</em>, 0) | 1 ≤ <em>x</em> ≤ 4} = 6 when <em>x</em> = 1

• <em>y</em> = 2   ==>   <em>f</em> (<em>x</em>, 2) = <em>x</em> ² - 8<em>x</em> + 5 = (<em>x</em> - 4)² - 11

Then

max{<em>f</em> (<em>x</em>, 2) | 1 ≤ <em>x</em> ≤ 4} = -2 when <em>x</em> = 1

min{<em>f</em> (<em>x</em>, 2) | 1 ≤ <em>x</em> ≤ 4} = -11 when <em>x</em> = 4

So to summarize, we found

max{<em>f(x, y)</em> | 1 ≤ <em>x</em> ≤ 4, 0 ≤ <em>y</em> ≤ 2} = 21 at (<em>x</em>, <em>y</em>) = (4, 0)

min{<em>f(x, y)</em> | 1 ≤ <em>x</em> ≤ 4, 0 ≤ <em>y</em> ≤ 2} = -11 at (<em>x</em>, <em>y</em>) = (4, 2)

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Answer:

Step-by-step explanation:

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