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2 years ago

5.2x-y=4.1 1.5x+y=6.7 how to solve this problem

Mathematics

2 answers:

MrRa [10]2 years ago

8 0

Answer: x=1.6

y=4.2

Step-by-step explanation:

5.2x - y = 4.1

1.5x + y = 6.7

we will add both equations and you see that the y is getting cancelled

6.7x=10.8

x=10.8/6.7= 1.6

now in any equation we will replace x with 1.6

5.2(1.6)-y=4.1

8.3-y=4.1

-y= 4.1-8.3

-y=-4.2 multiply by -1

y=4.2

sesenic [268]2 years ago

4 0

Answer:

x = 1.61 (Estimated)

y = 4.28 (Estimated)

Step-by-step explanation:

5.2x - y = 4.1

1.5x + y = 6.7

5.2x − y = 4.1

−y = −5.2x + 4.1

y = 5.2x − 4.1

Solve for x:

1.5x + y = 6.7

1.5x + 5.2x − 4.1 = 6.7

6.7x − 4.1 = 6.7

6.7x = 10.8

x = 1.61194

Solve for y:

y = 5.2x − 4.1

y = 5.2 × 1.61194 − 4.1

y = 4.28209

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Assuming that the population is normally distributed, construct a 9090% confidence interval for the population mean for each o

jasenka [17]

Given:

Set A: 1 4 4 4 5 5 5 8

Mean: 4.5

Standard dev: 1.9

Set B:

Mean: 4.5

Standard dev: 2.45

% = 90

Set A:

Standard Error, SE = s/ √n = 1.9/√8 = 0.67

Degrees of freedom = n - 1 = 8 -1 = 7

t- score = 1.89457861

Width of the confidence interval = t * SE = 1.89457861* 0.67 = 1.272685913

Lower Limit of the confidence interval = x-bar - width = 4.5 - 1.272685913 = 3.23

Upper Limit of the confidence interval = x-bar + width = 4.5 + 1.272685913 = 5.77

The 90% confidence interval is [3.23, 5.77]

Set B:

Standard Error, SE = s/ √n = 2.45/√8 = 0.87

Degrees of freedom = n - 1 = 8 -1 = 7

t- Score = 1.89457861

Width of the confidence interval = t * SE = 1.89457861* 0.87 = 1.641094994

Lower Limit of the confidence interval = x-bar - width = 4.5 - 1.641094994 = 2.86

Upper Limit of the confidence interval = x-bar + width = 4.5 + 1.641094994 = 6.14

The 90% confidence interval is [2.86, 6.14]

We can obviously see that sample B has more variation in the scores than sample A. The fact that the standard deviation is 2.45 for B and 1.9 for A). Therefore, they yield dissimilar confidence intervals even though they have the same mean and range.

6 0

3 years ago

Can someone help me with answer C? its the last one i need

Dmitry [639]

Using the monthly payment formula, it is found that her down payment should be of $1,419.

<h3>What is the monthly payment formula?</h3>

It is given by:

$A = P\frac{\frac{r}{12}\left(1 + \frac{r}{12}\right)^n}{\left(1 + \frac{r}{12}\right)^n - 1}$

In which:

P is the initial amount.

r is the interest rate.

n is the number of payments.

For this problem, the parameters are:

A = 250, r = 0.072, n = 72.

Hence:

r/12 = 0.072/12 = 0.006.

We solve for P to find the total amount of the monthly payments, hence:

$A = P\frac{\frac{r}{12}\left(1 + \frac{r}{12}\right)^n}{\left(1 + \frac{r}{12}\right)^n - 1}$

$P\frac{0.006(1.006)^{72}}{(1.006)^{72}-1} = 250$

0.0171452057P = 250

P = 250/0.0171452057

P = $14,581.

The total payment is of $16,000, hence her down payment should be of:

16000 - 14581 = $1,419.

More can be learned about the monthly payment formula at brainly.com/question/26476748

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4 0

2 years ago

HELP HELP HELP HELP HELP HELP

Anton [14]

hi hi hi fnjdjejw was your birthday and i’m is the time for you i’m i yw

6 0

3 years ago

If the quadrilateral below is a kite, find m

Debora [2.8K]

Answer:

49°

Step-by-step explanation:

7.5x-15+90=180

7.5x+75 =180

7.5x=105

x =14

so NPQ = ( 4×14 -7 ) = 49° this the answer

6 0

2 years ago

Read 2 more answers

The mean one-way commute to work in Chowchilla is 7 minutes. The standard deviation is 2.4 minutes, and the population is normal

adell [148]

Answer:

The answer is below

Step-by-step explanation:

Given that:

The mean (μ) one-way commute to work in Chowchilla is 7 minutes. The standard deviation (σ) is 2.4 minutes.

The z score is used to determine by how many standard deviations the raw score is above or below the mean. It is given by:

$z=\frac{x-\mu}{\sigma}$

a) For x < 2:

$z=\frac{x-\mu}{\sigma}=\frac{2-7}{2.4} =-2.08$

From normal distribution table, P(x < 2) = P(z < -2.08) = 0.0188 = 1.88%

b) For x = 2:

$z=\frac{x-\mu}{\sigma}=\frac{2-7}{2.4} =-2.08$

For x = 11:

$z=\frac{x-\mu}{\sigma}=\frac{11-7}{2.4} =1.67$

From normal distribution table, P(2 < x < 11) = P(-2.08 < z < 1.67 ) = P(z < 1.67) - P(z < -2.08) = 0.9525 - 0.0188 = 0.9337

c) For x = 11:

$z=\frac{x-\mu}{\sigma}=\frac{11-7}{2.4} =1.67$

From normal distribution table, P(x < 11) = P(z < 1.67) = 0.9525

d) For x = 2:

$z=\frac{x-\mu}{\sigma}=\frac{2-7}{2.4} =-2.08$

For x = 5:

$z=\frac{x-\mu}{\sigma}=\frac{5-7}{2.4} =-0.83$

From normal distribution table, P(2 < x < 5) = P(-2.08 < z < -0.83 ) = P(z < -0.83) - P(z < -2.08) = 0.2033- 0.0188 = 0.1845

e) For x = 5:

$z=\frac{x-\mu}{\sigma}=\frac{5-7}{2.4} =-0.83$

From normal distribution table, P(x < 5) = P(z < -0.83) = 0.2033

8 0

3 years ago