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Svetlanka [38]
2 years ago
4

A jacket for $75.00 cash, or $25.00 down and $4.75 per week for 11 weeks

Mathematics
1 answer:
Art [367]2 years ago
6 0

Answer:

It's wrong, the jacket cost 77.25 $, but if you pay 75.00$ cash (right away), you will earn a discount of 2.25

Step-by-step explanation:

The store offers the jacket's in 75.00 cash

but if you put 25.00 cash you have a proposal where you can pay 4.75 per week, for 11 weeks, that will be a total amount of 4.75*11= 52.25

If you put all together, you will pay an amount of 25+52.25=77.25.

That means that the store earns an extra of 2.25.

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Find a matrix P such that PTAP orthogonally diagonalizes A. Verify that PTAP gives the proper diagonal form. (Enter each matrix
krok68 [10]

Answer:

the P matrix you are looking for is P=(1/\sqrt{2}) · [[1 1 0 0],[1 -1 0 0],[0 0 1 1],[0 0 1 -1]]

Step-by-step explanation:

Answer:

For an orthogonal diagonalization of any matrix you have to:

1º) Find the matrix eigenvalues in a set order.

2º) Find the eigenvectors of each respective eigenvalues.

Tip: You can write the matrix A like A = P^{t} D P

3º) D is the diagonal matrix with each eigenvalue (in order) in the diagonal.

4º) Write P as the normalized eigenvectors in order (in columns).

Tip 2: Remember, P^{t}·P = I, so if A = P^{t} D P, then:

P A P^{t} = P  P^{t} D P P^{t} = I D I = D

So the P we are looking for is the P^{t} of the diagonalization.

Tip 3: In this case, A is a block matrix with null nondiagonal submatrixes, therefore its eigenvalues can be calculated by using the diagonal submatrixes. The problem is reduced to calculate the eigenvalues of A₁₁ = A ₂₂ = [[5 3],[3 5]]

Solving:

1º)the eigenvalues of A₁₁ are {8,2}, therefore the D matrix is \left[\begin{array}{cccc}8&0&0&0\\0&2&0&0\\0&0&8&0\\0&0&0&2\end{array}\right]

2º) the eigenvectors of A₁₁ are P₈= {[1 1]T} P₂= {[1 -1]T}, therefore normalizing the eigenvectors you obtain P = (1/\sqrt{2}) · [[1 1 0 0],[1 -1 0 0],[0 0 1 1],[0 0 1 -1]] (you can see that P =  P^{t} in this case).

As said in "tip 2": P A P^{t} = P  P^{t} D P P^{t} = I D I = D

So the P obtained is the one you are looking for.

4 0
2 years ago
A soccer ball is kicked into the air from the ground. If the ball reaches a maximum height of 25 ft and spends a total of 2.5 s
kondaur [170]
Hello,

g=9.81m/s²= 9.81/0.3048 =32.18.. ft/s² rounded to 32

h=-g/2*t²+vt+h0
if t=0,h=0 ==>h0=0

==>h=-16*t²+vt
50 ft in 2.5s ==> 25 ft in 1.25s

25=-16*1.25²+v*1.25
==>v=50/1.25=40

Equation h=-16t²+40*t
6 0
2 years ago
A standard weight known to weigh 10 grams. Some suspect bias in weights due to manufacturing process. To assess the accuracy of
notsponge [240]

Answer:

a) The 98% confidence interval for the mean weight is between 10.00409 grams and 10.00471 grams

b) 49 measurements are needed.

Step-by-step explanation:

Question a:

We have that to find our \alpha level, that is the subtraction of 1 by the confidence interval divided by 2. So:

\alpha = \frac{1 - 0.98}{2} = 0.01

Now, we have to find z in the Ztable as such z has a pvalue of 1 - \alpha.

That is z with a pvalue of 1 - 0.01 = 0.99, so Z = 2.327.

Now, find the margin of error M as such

M = z\frac{\sigma}{\sqrt{n}}

In which \sigma is the standard deviation of the population and n is the size of the sample.

M = 2.327\frac{0.0003}{\sqrt{5}} = 0.00031

The lower end of the interval is the sample mean subtracted by M. So it is 10.0044 - 0.00031 = 10.00409 grams

The upper end of the interval is the sample mean added to M. So it is 10 + 0.00031 = 10.00471 grams

The 98% confidence interval for the mean weight is between 10.00409 grams and 10.00471 grams.

(b) How many measurements must be averaged to get a margin of error of +/- 0.0001 with 98% confidence?

We have to find n for which M = 0.0001. So

M = z\frac{\sigma}{\sqrt{n}}

0.0001 = 2.327\frac{0.0003}{\sqrt{n}}

0.0001\sqrt{n} = 2.327*0.0003

\sqrt{n} = \frac{2.327*0.0003}{0.0001}

(\sqrt{n})^2 = (\frac{2.327*0.0003}{0.0001})^2

n = 48.73

Rounding up

49 measurements are needed.

7 0
2 years ago
Find the length of the median of a trapezoid if the length
guapka [62]

Answer:

20 cm

Step-by-step explanation:

We are given a trapezoid, where the length of shorter base or on of the parllel line is 16 cm and the length of other parallel side is 24 cm.

Let the two parallel sides be x and y that is x = 16 cm and y = 24 cm.

A median of a trapezoid is a line segment that divides the non parallel sides of a trapezoid equally or a line segment that passes through the mid points of non-parallel sides of a trapezoid.

The length of median of a trapezoid = \frac{\text{Sum of parallel sides}}{2} = \frac{16+24}{2} = 20 cm.

Thus, the length of median of trapezoid is 20 cm.

8 0
2 years ago
The United States Bureau of Labor Statistics (BLS) conducts the Quarterly Census of Employment and Wages (QCEW) and reports a va
Firlakuza [10]

Answer:

1

  The  probability is  P(\= X < 33) = 0.8531

2

  The  probability is  P(\= X > 30) = 0.3520

Step-by-step explanation:

From the  question we are told that

     The population mean is  \mu =  28.29

      The standard deviation is \sigma  =  33.493

       The sample size is  n  = 56

Generally the standard error for the  sample  mean (\= x ) is mathematically evaluated as

        \sigma _{\=x} =  \frac{\sigma}{\sqrt{n} }

substituting values  

       \sigma _{\=x} =  \frac{33.493}{\sqrt{56} }

      \sigma _{\=x} = 4.48

Apply central limit theorem[CLT] we have  that

        P(\= X < 33) =  [z <  \frac{33 -  \mu }{\sigma_{\= x}} ]

substituting values

       P(\= X < 33) =  [z <  \frac{33 -  28.29 }{4.48} ]

       P(\= X < 33) =  [z <  1.05 ]

From the z-table  we have that  

       P(\= X < 33) = 0.8531

For the second question

    Apply central limit theorem[CLT] we have  that

    P(\= X > 30 ) =  [z >  \frac{30 -  \mu }{\sigma_{\= x}} ]

substituting values

   P(\= X < 33) =  [z >  \frac{30 -  28.29 }{4.48} ]

From the z-table  we have that  

     P(\= X < 30) = 0.6480

Thus  

     P(\= X > 30) = 1- P(\= X < 30) = 1- 0.6480

     P(\= X > 30) = 0.3520

3 0
2 years ago
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