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3 years ago

What is the mean of: 3.7, 5, 9.2, 4, 6.1, 5, 2.6, 4, 5.2, 5?

Mathematics

1 answer:

mart [117]3 years ago

6 0

Answer:

4.98

Step-by-step explanation:

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Given that justin is collecting data on reaction time, what type of data is he working with?.

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Answer:

Given that Justin is collecting data on reaction time, what type of data is he working with? Reaction time is continuous quantitative data because it is obtained by measuring and is not limited to a certain set of numbers.

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2 years ago

Half of your baseball card collection got wet and was ruined. You bought 9 cards to replace some that were lost. How many did yo

PtichkaEL [24]

solution:

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6 0

3 years ago

1.Given the function h(t) = t³ - 2t, find h(-5).

Svetllana [295]

Answer:

-65

Step-by-step explanation:

h(t) = (-5)^3 -2(-5)

h(t) = -75 +10

h(t) = -65

7 0

1 year ago

find the centre and radius of the following Cycles 9 x square + 9 y square +27 x + 12 y + 19 equals 0

Citrus2011 [14]

Answer:

Radius: $r =\frac{\sqrt {21}}{6}$

$Center = (-\frac{3}{2}, -\frac{2}{3})$

Step-by-step explanation:

Given

$9x^2 + 9y^2 + 27x + 12y + 19 = 0$

Solving (a): The radius of the circle

First, we express the equation as:

$(x - h)^2 + (y - k)^2 = r^2$

Where

$r = radius$

$(h,k) =center$

So, we have:

$9x^2 + 9y^2 + 27x + 12y + 19 = 0$

Divide through by 9

$x^2 + y^2 + 3x + \frac{12}{9}y + \frac{19}{9} = 0$

Rewrite as:

$x^2 + 3x + y^2+ \frac{12}{9}y =- \frac{19}{9}$

Group the expression into 2

$[x^2 + 3x] + [y^2+ \frac{12}{9}y] =- \frac{19}{9}$

$[x^2 + 3x] + [y^2+ \frac{4}{3}y] =- \frac{19}{9}$

Next, we complete the square on each group.

For $[x^2 + 3x]$

1: Divide the $coefficient\ of\ x\ by\ 2$

2: Take the $square\ of\ the\ division$

3: Add this $square\ to\ both\ sides\ of\ the\ equation.$

So, we have:

$[x^2 + 3x] + [y^2+ \frac{4}{3}y] =- \frac{19}{9}$

$[x^2 + 3x + (\frac{3}{2})^2] + [y^2+ \frac{4}{3}y] =- \frac{19}{9}+ (\frac{3}{2})^2$

Factorize

$[x + \frac{3}{2}]^2+ [y^2+ \frac{4}{3}y] =- \frac{19}{9}+ (\frac{3}{2})^2$

Apply the same to y

$[x + \frac{3}{2}]^2+ [y^2+ \frac{4}{3}y +(\frac{4}{6})^2 ] =- \frac{19}{9}+ (\frac{3}{2})^2 +(\frac{4}{6})^2$

$[x + \frac{3}{2}]^2+ [y +\frac{4}{6}]^2 =- \frac{19}{9}+ (\frac{3}{2})^2 +(\frac{4}{6})^2$

$[x + \frac{3}{2}]^2+ [y +\frac{4}{6}]^2 =- \frac{19}{9}+ \frac{9}{4} +\frac{16}{36}$

Add the fractions

$[x + \frac{3}{2}]^2+ [y +\frac{4}{6}]^2 =\frac{-19 * 4 + 9 * 9 + 16 * 1}{36}$

$[x + \frac{3}{2}]^2+ [y +\frac{4}{6}]^2 =\frac{21}{36}$

$[x + \frac{3}{2}]^2+ [y +\frac{4}{6}]^2 =\frac{7}{12}$

$[x + \frac{3}{2}]^2+ [y +\frac{2}{3}]^2 =\frac{7}{12}$

Recall that:

$(x - h)^2 + (y - k)^2 = r^2$

By comparison:

$r^2 =\frac{7}{12}$

Take square roots of both sides

$r =\sqrt{\frac{7}{12}}$

Split

$r =\frac{\sqrt 7}{\sqrt 12}$

Rationalize

$r =\frac{\sqrt 7*\sqrt 12}{\sqrt 12*\sqrt 12}$

$r =\frac{\sqrt {84}}{12}$

$r =\frac{\sqrt {4*21}}{12}$

$r =\frac{2\sqrt {21}}{12}$

$r =\frac{\sqrt {21}}{6}$

Solving (b): The center

Recall that:

$(x - h)^2 + (y - k)^2 = r^2$

Where

$r = radius$

$(h,k) =center$

From:

$[x + \frac{3}{2}]^2+ [y +\frac{2}{3}]^2 =\frac{7}{12}$

$-h = \frac{3}{2}$ and $-k = \frac{2}{3}$

Solve for h and k

$h = -\frac{3}{2}$ and $k = -\frac{2}{3}$

Hence, the center is:

$Center = (-\frac{3}{2}, -\frac{2}{3})$

6 0

2 years ago

PLEASE ALSO EXPLAIN HOW TO SOLVE THIS. 20 POINTS

eimsori [14]

B.) Hibiscus Street

The key thing to remember is the slope of the lines. The equation for a line in slope intercept form is
y = ax + b
where
a = slope
b = y intercept

So the equation for Oak Street is y = 2/3x - 7. So it's slope is 2/3. And any street that has the same slope will be given a tree name. And any street that's perpendicular will be given a flower name. You can determine is a line is perpendicular if it has a slope that's the negative reciprocal. So a street that's perpendicular to Oak street will have a slope of -3/2.

Now you've just been given the equation to a new street that's y = -3/2x - 2. Since the slope is -3/2 and Oak street has a slope of 2/3, the new street is perpendicular to Oak street. And given the naming scheme, that means that the new street will have the name of a flower. So let's look at the available street names and pick a flower.
<span>
A.) Weeping Willow Street
* Nope a Weeping Willow is a tree. So this name won't work.

B.) Hibiscus Street
* Yes. A Hibiscus is a flower, so this name is suitable.

C.) Oak Street
* Nope. Not only is this a tree instead of a flower, but there's already an Oak street. So bad choice.

D.) Panther Street
* Nope, this is an animal, not a flower. Bad choice.

</span>

6 0

3 years ago