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3 years ago

What is the solution for x+y>2 2x-y>1

Mathematics

1 answer:

blagie [28]3 years ago

3 0

Answer:

x>1 and y>1

Step-by-step explanation:

x+y>2....... (i)

2x-y>1...... (ii)

Adding equations (i) and (ii),

x+y+2x-y>2+1

3x>3

x>3/3

x>1.

Putting the value of x in equation (i),

1+y>2

y>2-1

y>1.

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Health insurers are beginning to offer telemedicine services online that replace the common office visit. A company provides a v

Veronika [31]

Answer:

$\text {CI} = (60.54, \: 81.46)\\\\$

Therefore, we are 95% confident that actual mean savings for a televisit to the doctor is within the interval of ($60.54 to $81.46)

Step-by-step explanation:

Let us find out the mean savings for a televisit to the doctor from the given data.

Using Excel,

=AVERAGE(number1, number2,....)

The mean is found to be

$\bar{x} = \$71$

Let us find out the standard deviation of savings for a televisit to the doctor from the given data.

Using Excel,

=STDEV(number1, number2,....)

The standard deviation is found to be

$s = \$ 22.35$

The confidence interval is given by

$\text {confidence interval} = \bar{x} \pm MoE\\\\$

Where the margin of error is given by

$$ MoE = t_{\alpha/2}(\frac{s}{\sqrt{n} } ) $ \\\\$

Where n is the sample of 20 online doctor visits, s is the sample standard deviation and $t_{\alpha/2}$ is the t-score corresponding to a 95% confidence level.

The t-score is given by is

Significance level = α = 1 - 0.95 = 0.05/2 = 0.025

Degree of freedom = n - 1 = 20 - 1 = 19

From the t-table at α = 0.025 and DoF = 19

t-score = 2.093

So, the margin of error is

$MoE = t_{\alpha/2}(\frac{s}{\sqrt{n} } ) \\\\MoE = 2.093\cdot \frac{22.35}{\sqrt{20} } \\\\MoE = 2.093\cdot 4.997\\\\MoE = 10.46\\\\$

So the required 95% confidence interval is

$\text {CI} = \bar{x} \pm MoE\\\\\text {CI} = 71 \pm 10.46\\\\\text {CI} = 71 - 10.46, \: 71 + 10.46\\\\\text {CI} = (60.54, \: 81.46)\\\\$

Therefore, we are 95% confident that actual mean savings for a televisit to the doctor is within the interval of ($60.54 to $81.46)

7 0

3 years ago

The frequency of the musical note C4 is about 261.63 Hz. What is the frequency of the note a perfect fifth below C4?

kolbaska11 [484]

Answer:

The frequency of the note a perfect fifth below C4 is;

B- 174.42 Hz

Step-by-step explanation:

Here we note that to get the "perfect fifth" of a musical note we have to play a not that is either 1.5 above or 1.5 below the note to which we reference. Therefore to get the frequency of the note a perfect fifth below C4 which is about 261.63 Hz, we have

1.5 × Frequency of note Y = Frequency of C4

1.5 × Y = 261.63

Therefore, Y = 261.63/1.5 = 174.42 Hz.

6 0

3 years ago

What's one way we can check our answer to 2 x 10 = 20

julsineya [31]

You can do 20/2=10 This helps because it is just doing the reverse so you know if you get it both ways that it is correct. Medal please?

8 0

3 years ago

Read 2 more answers

Fatima earns $150 in 4 hours. At this rate, how many hours will it take her to earn $675?

Vedmedyk [2.9K]

Answer:

$150 = 4 hours,

So, in a ratio, 4 : 150

=> 150 * 4.5 : 4 * 4.5

=> 675 : 18 hours

Therefore to earn $675 at the rate of $150 for 4 hours, Fatima will need to work for 18 hours to earn $675

If my answer helped, kindly mark me as the brainliest!!

Thank You!!

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3 years ago

Find the sum of 7.7, 0.94, and 13

poizon [28]

Answer:

21.64

Step-by-step explanation:

7.7+0.94+13=21.64

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3 years ago