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Ivanshal [37]
3 years ago
10

-3 = -11 # -10 H + 05 > -15 -5 10 15 Stuck? Review related articles/videos or use a hint.​

Mathematics
1 answer:
ElenaW [278]3 years ago
4 0

weooooooooooooooooooooooooooooooooooooo9ooooooooooow

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HEY IF YOURE SEEING THIS IM IN DIRE NEED OF YOU SOLVING NUMBER 4!!
gavmur [86]

Answer:

(1, -3) (2, -1) (3, 1)

Step-by-step explanation:

if seconds=x and degrees celsius=y, the points are (1, -3) (2, -1) (3, 1). to graph, plot the points on a graph. (ordered pairs are written as x,y)

8 0
2 years ago
English 3 hhhhhhhhhheeeeeeelllllllllllllllllppppppppppp me
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Answer:

the answer is: A

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3 years ago
Express 6^1/3 in simplest radical form.
seropon [69]

<em>Greetings from Brasil</em>

From radiciation properties:

\large{A^{\frac{P}{Q}}=\sqrt[Q]{A^P}}

bringing to our problem

\large{6^{\frac{1}{3}}=\sqrt[3]{6^1}}

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4 years ago
In measuring the center of the data from a skewed distribution, the median would be preferred over the mean for most purposes be
icang [17]
<span>The median would be preferred over the mean in such scenarios because the median will lessen the impact of the outliers that fall within the "tail" of the skew. Therefore, if a curve is normally distributed, that is to say that data is normally distributed, there will be two tails, each with approximately equal proportions of outliers. Outliers in this case being more extreme numbers, and are based on your determination depending on how you are using the data. If data is skewed there is one tail, and therefore it may be an inaccurate measure of central tendency if you use the mean of the numbers. Thinking of this visually. In positively skewed data where there is a "tail" towards the right and a "peak" towards the left, the median will be placed more in the "peak", whereas the mean will be placed more towards the "tail", making it a poorer measure of central tendency, or the center of the data.</span>
5 0
3 years ago
You are riding a roller coaster when a shoe falls off your foot. The function y=300−16t2 represents the height y (in feet) of th
Juli2301 [7.4K]

Answer:

Time to land on 11ft is 4.25 seconds

Step-by-step explanation:

Given

y = 300 - 16t^2

Required

Solve for t when height is 11ft

This implies that y = 11

So: substitute 11 for y in y = 300 - 16t^2

11 = 300 - 16t^2

Collect Like Terms

16t^2 = 300 - 11

16t^2 = 289

Express both sides as squares

4^2 * t^2 = 17^2

(4t)^2 = 17^2

Take square roots of both sides (ignore negative)

4t = 17

Solve for t

t = \frac{17}{4}

t = 4.25

Hence, time to land on 11ft is 4.25 seconds

8 0
3 years ago
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