PLEASE HELP IT'S URGENT

Solve the system of two equations 2y=8 and 15y=2x-4

damaskus [11]

To find y, divide the first equation by 2.
y = 4

To find x, use that value for y in the second equation. Add 4, then divide by 2.
15·4 = 2x -4
60 +4 = 2x
64/2 = x = 32

The solution is (x, y) = (32, 4).

8 0

2 years ago

Cosa/2-tana +sina/1-cota=cosa+sina

Inessa05 [86]

Answer:

Cosa/2-tana+sina

Step-by-step explanation:

Cosatana

7 0

11 months ago

confirm or disprove ernest's work by selecting two different points and applying the slope formula. be sure to identify the poin

yan [13]

Answer:

Part a) Ernest's calculation was correct

Part b) Ernest's calculation was correct

Step-by-step explanation:

The picture of the question in the attached figure

<u><em>The complete question is</em></u>

Ernest calculated the slope of the graph during the uphill climb to be 10. His work is shown below.

Points: (60.900). (15, 450)

Slope : 900-450/60-15=450/45=10

Part a) Determine whether his calculation was correct or incorrect, and then explain why.

Part b) Confirm or disprove Ernest's work by selecting two different points and applying the slope formula. be sure to identify the points you chose

Part a)

we know that

The formula to calculate the slope between two points is equal to

$m=\frac{y2-y1}{x2-x1}$

we have the points

A(60,900),B(15,450)

Remember that

$m_A_B=m_B_A$

step 1

Find $m_A_B$

substitute in the formula

$m_A_B=\frac{450-900}{15-60}\\\\m_A_B=\frac{-450}{-45}=10$

step 2

Find $m_B_A$

substitute in the formula

$m_B_A=\frac{900-450}{60-15}\\\\m_B_A=\frac{450}{45}=10$

therefore

Ernest's calculation was correct, because $m_A_B=m_B_A$

Part b) selecting two different points

we take the points

(0,300) and (50,800) -----> see the attached figure

substitute in the formula of slope

$m=\frac{800-300}{50-0}$

$m=\frac{500}{50}$

$m=10$

therefore

Ernest's calculation was correct

8 0

2 years ago

If you are paid every 3 months, or 4 times a year, what is that called?

ollegr [7]

Thank you! This helped a lot!!! :)

5 0

1 year ago

A sample of blood pressure measurements is taken for a group of adults, and those values (mm Hg) are listed below. The values

Slav-nsk [51]

Answer:

Systolic on right

$\hat{CV} =\frac{18.68}{127.5}=0.147$

Systolic on left

$\hat{CV} =\frac{12.65}{74.2}=0.170$

So for this case we have more variation for the data of systolic on left compared to the data systolic on right but the difference is not big since 0.170-0.147 = 0.023.

Step-by-step explanation:

Assuming the following data:

Systolic (#'s on right) Diastolic (#'s on left)

117; 80

126; 77

158; 76

96; 51

157; 90

122; 89

116; 60

134; 64

127; 72

122; 83

The coefficient of variation is defined as " a statistical measure of the dispersion of data points in a data series around the mean" and is defined as:

$CV= \frac{\sigma}{\mu}$

And the best estimator is $\hat {CV} =\frac{s}{\bar x}$

Systolic on right

We can calculate the mean and deviation with the following formulas:

[te]\bar x = \frac{\sum_{i=1}^n X_i}{n}[/tex]

$s= \frac{\sum_{i=1}^n (x_i -\bar X)^2}{n-1}$

For this case we have the following values:

$\bar x = 127.5, s= 18.68$

So then the coeffcient of variation is given by:

$\hat{CV} =\frac{18.68}{127.5}=0.147$

Systolic on left

For this case we have the following values:

$\bar x = 74.2 s= 12.65$

So then the coeffcient of variation is given by:

$\hat{CV} =\frac{12.65}{74.2}=0.170$

So for this case we have more variation for the data of systolic on left compared to the data systolic on right but the difference is not big since 0.170-0.147 = 0.023.

4 0

2 years ago