If the circle above is dilated using a scale factor of 5, what will be the radius of the resulting circle?

MrMuchimi

Anyone sees the problem? I can’t see it.

5 0

3 years ago

A beaker was filled with 10 ounces of water before starting a science experiment. After the experiment, 6 ounces of water remain

Allisa [31]

There were 4 fewer ounces of water after the experiment

3 0

3 years ago

Read 2 more answers

Three consecutive odf integers have a sum of 39

Anestetic [448]

11, 13, 15
...............

7 0

4 years ago

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The oil prices for 2014, rounded to the nearest dollar, were: 95, 101, 101, 102, 102, 106, 104, 97, 93, 84, 76, 59. what is the

kipiarov [429]

Interquartile Range (IQR) = $Q_3-Q_1$ , with $Q_3$ as the upper quartile and $Q_1$ as the lower quartile.

Firstly, rearrange the data so that it's in ascending order: $\{59,76,84,93,95,97,101,101,102,102,104,106\}$

Next, find the median:

$\{59,76,84,93,95,\boxed{97,101,}101,102,102,104,106\}\\\\\frac{97+101}{2}\\\\\frac{198}{2}\\\\99$

Now to find the lower quartile, find the "median" of the data set that's to the left of 99:

$\{\overbrace{59,76,84,93,95,97}^{\textsf{to the left of the median}},101,101,102,102,104,106\}\\\\\{\overbrace{59,76,\boxed{84,93}95,97}^{\textsf{to the left of the median}},101,101,102,102,104,106\}\\\\\frac{84+93}{2}\\\\\frac{177}{2}\\\\88.5$

Now to find the upper quartile, it's the similar process as finding the lower quartile, except that you are finding the "median" of the data set to the right of 99:

$\{59,76,84,93,95,97,\overbrace{101,101,102,102,104,106}^{\textsf{to the right of the median}}\}\\\\\{59,76,84,93,95,97,\overbrace{101,101,\boxed{102,102} 104,106}^{\textsf{to the right of the median}}\}\\\\\frac{102+102}{2}\\\\\frac{204}{2}\\\\102$

Now that we have the upper and lower quartile, subtract them:

$102-88.5=13.5$

<u>In short, the IQR of this data set is 13.5.</u>

8 0

3 years ago

Graph the lines by making a table. a. y = 2x = 4

AnnZ [28]

Answer:

Given y-2x=4 →y=2x+4

Table of Solutions

xy = 2x + 4(x, y) * P * o * i * n * t

0 y = 2(0) + 4 = 4 (0,4)A(0,4)

2 y = 2(-2) + 4 = 0 (-2, 0) B(-2, 0)

1y = 2(1) + 4 = 6(1, 6) C(1, 6)

Plotting the points A, B and C on the graph paper and joint them to get the straight line BC as shown in graph sheet. This line is the required graph of the equation y - 2x = 4 .

(i) Plot the point (2, 8) on the graph paper. From the graph it is clear that the point (2, 8) lies on the line.

Checking algebraically: On substituting (2, 8) in the

given equation, we get LHS y-2x=8-2×2=8-4-4 RHS, So (2, 8) is a solution

(ii) Plot the point (4, 2) on the graph paper. You fin that (4, 2) does not lie on the line.

Checking algebraically: By substituting (4, 2) in the given equation we have

LHS =y-2x=2-2×4-2-8--6

=RHS, So (4, 2) is not a solution.

Step-by-step explanation:

HOPE IT HELPS

7 0

3 years ago