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Firlakuza [10]
2 years ago
10

The following data shows the low temperature in Vancouver, in degrees Celsius, for two weeks in January one year. 8.90, 8.35, 8.

40, 8.40, 8.50, 8.20, 8.50, 8.32, 8.50, 8.50, 8.60, 8.30, 8.10, 8.65 Which box plot best represents this data?

Mathematics
1 answer:
frutty [35]2 years ago
5 0
There is no option of the box plots, so I have created a version that would represent this data.

To make the box plot you will need the lower extreme, lower quartile, median, upper quartile, and upper extreme.

Please see the attached picture.

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In ΔKLM, m = 59 cm, ∠L=111° and ∠M=60°. Find the length of l, to the nearest centimeter.
g100num [7]

Answer:

64 (or 63.6)

Step-by-step explanation:

You can figure out the <K by adding the first two angles (111+60) and getting 171 (subtracting that from 180 gives you a 9-degree angle for <K

Given three angles and 1 side, you can calculate the sides as follows:

k=m*sin (K) / sin (M) = 10.65

l=m*sin (L) / sin (M) = 63.60234

Since you're solving for l, the steps would be:

l=59 * sin (111) / sin (60)

l=59 * (0.934) / (0.866)

l=55.106 / 0.866

l=63.63

Hope this helps!

4 0
2 years ago
A meteorologist is studying the monthly rainfall in a section of the Brazilian rainforest. She recorded the monthly rainfall, in
aleksandr82 [10.1K]

Answer:

12.405

Step-by-step explanation:

3 0
2 years ago
Read 2 more answers
Which statements accurately describe the function f(x) = 3(16) Superscript three-fourths x? Select three options.
zzz [600]

Answer:

The initial value is 3.

The range is y greater than 0.

The simplified base is 8.

Step-by-step explanation:

The given function is f(x) = 3(16)^{\frac{3}{4}x } ......... (1)

Therefore, the initial value of the function at x = 0 is f(0) = 3(16)^{0} = 3  

Now, the domain can be any real value, since for all real value of x, y exists.

But, for no value of x the function has value < 0.

Therefore, y greater than 0 is the range of the function f(x).

Now, simplifying the equation (1) we will have  

f(x) = 3(16)^{\frac{3}{4}x } = 3(16^{\frac{3}{4} } )^{x} = 3(8)^{x}

Therefore, the simplified base is 8. (Answer)

7 0
2 years ago
Read 2 more answers
Consider the following equation, bh+hr=25 when solving for r
Komok [63]
Solve for r.

You want to get r by itself on one side on the equal sign.

bh + hr = 25

Subtract bh from both sides.

hr = 25 - bh

Divide h on both sides.

r = 25 - bh / h

The two h's cancel each other out.

r = 25 - b

Hope this helps!
4 0
2 years ago
Read 2 more answers
Use the position function s(t) = −16t² + 400, which gives the height (in feet) of an object that has fallen for t seconds from a
skad [1K]

Answer:

160m/s

Step-by-step explanation:

The object can hit the ground when t = a; meaning that s(a) = s(t) = 0

So, 0 = -16a² + 400

16a² = 400

a² = 25

a = √25

a = 5 (positive 5 only because that's the only physical solution)

The instantaneous velocity is

v(a) = lim(t->a) [s(t) - s(a)]/[t-a)

Where s(t) = -16t² + 400

and s(a) = -16a² + 400

v(a) = Lim(t->a) [-16t² + 400 + 16a² - 400]/(t-a)

v(a) = Lim(t->a) (-16t² + 16a²)/(t-a)

v(a) = lim (t->a) -16(t² - a²)(t-a)

v(a) = -16lim t->a (t²-a²)(t-a)

v(a) = -16lim t->a (t-a)(t+a)/(t-a)

v(a) = -16lim t->a (t+a)

But a = t

So, we have

v(a) = -16lim t->a 2a

v(a) = -32lim t->a (a)

v(a) = -32 * 5

v(a) = -160

Velocity = 160m/s

7 0
2 years ago
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