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3 years ago

Write the next five numbers in each sequence,

Mathematics

1 answer:

DIA [1.3K]3 years ago

3 0

Answer:

SOLUTION

First, we write the augmented matrix.

⎡

⎢

⎣

−

⎤

⎥

⎦

Next, we perform row operations to obtain row-echelon form.

−

→

⎡

⎢

⎣

−

⎤

⎥

⎦

−

→

⎡

⎢

⎣

−

⎤

⎥

⎦

The easiest way to obtain a 1 in row 2 of column 1 is to interchange \displaystyle {R}_{2}R

and \displaystyle {R}_{3}R

Interchange

and

→

⎡

⎢

⎣

−

⎤

⎥

⎦

Then

−

→

⎡

⎢

⎣

−

⎤

⎥

⎦

−

→

⎡

⎢

⎣

−

⎤

⎥

⎦

The last matrix represents the equivalent system.

−

Using back-substitution, we obtain the solution as \displaystyle \left(4,-3,1\right)(4,−3,1).

Step-by-step explanation:

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4 years ago

Divide 17/18 by 2/4 as a reduced fraction

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Answer:

17/9

Step-by-step explanation:

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3 years ago

The Weiland Computer Corporation is trying to choose between the following mutually exclusive design projects, P1 and P2:

uranmaximum [27]

Answer:

a profitability index P1=1.27

profitability index P2= 1.41

b NPV P1 = $14145.01

NPV P2 = $6630.3

c YES ANSWERS ARE DIFFERENT due to fact that cash flows in P1 is higher than in P2

Step-by-step explanation:

profitability index or = present value of an investment cash flows=67145.01

benefit cost ratio initial cost 53000

For project 1 or P1

cost of capital 10 % = 1.27

initial investment = $53,000 since year 0

Year Cash flows (P1)$ present value of future cash flows PV

1 27,000 24545.46

2 27,000 22314.05

3 27,000 20285.50

Total = 67145.01

use the formula of present value of future cash flows = C/(1+i)ⁿ

C = cash = 27000

i = interest = 10% = 10/ 100 = 0.1

n = year = 1

year 1 = 27000/(1+0.1)¹ = 24545.46

year 2 = 27000/(1.1)² = 22314.05 note n = 2

year 3 = 27000(1.1)³ = 20285.50

Profitability index = 1.27 > 1 thus it should be accepted

profitability index or = present value of an investment cash flows=22630.30

benefit cost ratio initial cost 16000

For project 1 or P1

cost of capital 10 % = 1.41

initial investment = $16,000 since year 0

Year Cash flows (P2)$ present value of future cash flows PV

1 9,100 8272.73

2 9,100 7520.6

3 9,100 6836.97

Total = 22630.30

use the formula of present value of future cash flows = C/(1+i)ⁿ

C = cash = 9100

i = interest = 10% = 10/ 100 = 0.1

n = year = 1

year 1 = 9100/(1+0.1)¹ = 8272.73

year 2 = 9100/(1.1)² = 7520.6 note n = 2

year 3 = 9100(1.1)³ = 6836.97

Profitability index = 1.41 > 1 thus it should be accepted

Profitability index of P1 = 1.27 AND P2 1.41 SO Weiland Computer Corporation SHOULD TAKE P2

b Net Present Value (NPV) decision rule =∑ pv - initial investment required

P1 = 67145.01 - 53,000 = $14145.01

P2 = 22630.30 - 16000 = $6630.3

Weiland Computer should take P1 since P1 > P2 that is $14145.01 >$6630.3

C YES ANSWERS ARE DIFFERENT due to fact that cash flows in P1 is higher than in P2

6 0

3 years ago

20 liters of mixture contain milk nad water in the ratio 5:3 of 4 liters of the mixture are replaced by 4 liters of milk find th

pantera1 [17]

Answer:

7:3

Step-by-step explanation:

5 + 3 = 8

The ratio is

5 milk : 3 water : 8 total

Milk is 5/8 of the total.

Water is 3/8 of the total.

The 20-liter mixture contains:

5/8 * 20 = 12.5 liters of milk, and

3/8 * 20 = 7.5 liters of water

4 liters of the mixture contain:

5/8 * 4 = 2.5 liters of milk, and

3/8 * 4 = 1.5 liter of water

When you remove 4 liters of the mixture from 20 liters of the mixture, you end up with

12.5 L - 2.5 L = 10 L milk, and

7.5 L - 1.5 L = 6 L water

Now you add 4 liters of milk. Now you have

10 L + 4 L = 14 L milk

6 L water

The new ratio of milk to water is 14:6 = 7:3

7 0

3 years ago

Read 2 more answers

Refer to the data in the logic circuit waferexample on page 298 ( ẋ=0.16, s²=.000314,n=15). Should the industrial engineer accep

muminat

Answer:

a) If we compare the p value and a significance level assumed $\alpha=0.01$ we see that $p_v$ so we can reject the null hypothesis, and the the actual true mean is significantly lower than 0.18 ounces.

b) The rejection region zone should be: t <-2.62

c) The rejection region in terms of the mean is: $\bar x$ <0.168

Step-by-step explanation:

Data given and notation

$\bar X=0.16$ represent sample mean

$s=\sqrt{0.000314}=0.0177$ represent the standard deviation for the sample

$n=15$ sample size

$\mu_o =0.18$ represent the value that we want to test

$\alpha=0.01$ 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 determine if the mean is less than 0.18 ounces, the system of hypothesis would be:

Null hypothesis: $\mu \geq 0.18$

Alternative hypothesis: $\mu < 0.18$

We don't know the population deviation, so for this case 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:

$t=\frac{0.16-0.18}{\frac{0.0177}{\sqrt{15}}}=-4.37$

Calculate the P-value

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

$p_v =P(Z$

Conclusion

If we compare the p value and a significance level assumed $\alpha=0.01$ we see that $p_v$ so we can reject the null hypothesis, and the the actual true mean is significantly lower than 0.18 ounces.

Should the industrial engineer accept or reject the null hypothesis that μ>= 0.18 ounce?

We reject the null hypothesis.

Rejection region in terms of t: t <

On this case we need to find first the degrees of freedom given by:

$df=n-1=15-1=14$

Now since the test it's one left tailed test we need a value on the t distribution with 14 degrees of freedom such that we have 0.01 of the area on the left and 0.99 of the area on the right. For this case we can use the following excel code:

"=T.INV(0.01,14)" and we got that the rejection region zone should be: t <-2.62

Rejection region in terms of ẋ: ẋ < .

We can use th critical value founded on the last part and we can use this formula similar to the z score.

$t=\frac{\bar X -\mu}{\frac{s}{\sqrt{n}}}$

And we can solve for $\bar X$ like this:

$\bar X =\mu + t*\frac{s}{\sqrt{n}} =0.18-2.62*\frac{0.0177}{\sqrt{15}}=0.168$

And the rejection region in terms of the mean is: $\bar x$ <0.168

5 0

3 years ago