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Bumek [7]
3 years ago
15

Which type of correlation is suggested by the scatter plot

Mathematics
1 answer:
kondor19780726 [428]3 years ago
4 0

First one is negative correlation, second one is no correlation. You're welcome.


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Find the circumference of a sphere's great circle with a radius of 4 m.
Ulleksa [173]
This is the concept of areas and circumference of the solid figures;
The circumference of the cylinder whose radius is 4 m will be given by;
C=2πr
C=2*π*4
C=8π m=25.133 m
3 0
3 years ago
The clownfish kay wanted to buy for $12 last week has been marked up 8%. What is the new price of the fish?
Crank
The markup is 8%, meaning it costs 8% more.
The starting price is 100%.
Therefore, the markup is 108% of $12.
If we multiply 12 by 108% (1.08), we will get our answer of $12.96
4 0
3 years ago
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A square sheet of paper measures 42 centimeters on each side. What is the length of the diagonal of this​ paper?
Illusion [34]
59.4cm
Use Pythagorean theorem to find the diagonal (^ means exponent)
a^2 + b^2 = c^2
42^2 + 42^2= c^2
1764+1764=c^2
3528=c^2
square root both sides
c= 59.39696962
round and you get 59.4 cm
5 0
4 years ago
On a linear X temperature scale, water freezes at −115.0°X and boils at 325.0°X. On a linear Y temperature scale, water freezes
belka [17]

Answer:

The current temperature on the X scale is 1150 °X.

Step-by-step explanation:

Let is determine first the ratio of change in X linear temperature scale to change in Y linear temperature scale:

r = \frac{\Delta T_{X}}{\Delta T_{Y}}

r = \frac{325\,^{\circ}X-(-115\,^{\circ}X)}{-25\,^{\circ}Y - (-65.00\,^{\circ}Y)}

r = 11\,\frac{^{\circ}X}{^{\circ}Y}

The difference between current temperature in Y linear scale with respect to freezing point is:

\Delta T_{Y} = 50\,^{\circ}Y - (-65\,^{\circ}Y)

\Delta T_{Y} = 115\,^{\circ}Y

The change in X linear scale is:

\Delta T_{X} = r\cdot \Delta T_{Y}

\Delta T_{X} = \left(11\,\frac{^{\circ}X}{^{\circ}Y} \right)\cdot (115\,^{\circ}Y)

\Delta T_{X} = 1265\,^{\circ}X

Lastly, the current temperature on the X scale is:

T_{X} = -115\,^{\circ}X + 1265\,^{\circ}X

T_{X} = 1150\,^{\circ}X

The current temperature on the X scale is 1150 °X.

5 0
3 years ago
As a promotion, a clothing store draws the name of one of its customers each week. The prize is a coupon for the store. If the w
kompoz [17]

The given formula is f(x) = 20(1.2)^x

The formula is the starting amount multiplied by 1 + the percentage raised to the number of weeks.


A) the percent increase is 20% ( 1.2 in the formula is 1 +20% as a decimal)

B) the original amount is $20

C) for 2 weeks, replace x with 2 and solve:

20(1.2)^2

20(1.44) = $28.80

After 2 weeks the coupon is $28.80

D)  To solve for the number of weeks (x) set the equation equal to $100:

100 = 20(1.2)^x

Divide both sides by 20:

5 = 1.2^x

Take the natural logarithm of both sides:

ln(5) = ln(1.2^x)

Use the logarithm rule to remove the exponent:

ln(5) = x ln(1.2)

Divide both sides by ln(1.2)

x = ln(5) / ln(1.2)

Divide:

X = 8.83

At 8.83 weeks the coupon would be $100, so after 9 weeks the coupon would be greater than $100

The answer is 9 weeks.

4 0
3 years ago
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