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Tanya [424]
2 years ago
8

PLEASE HELP ME QUICKKK, FIRST CORRECT PERSON GETS BRAINLIEST​

Mathematics
2 answers:
Sedaia [141]2 years ago
8 0

The percent decrease: 25%

The percent decrease: 33.33..%

So, B is the correct statement

inna [77]2 years ago
7 0
C.)

Explanation: the percentages decreased was 25% while the percentage increased from 60 was 33%. From this 25< 33 so C. Would be best bet!
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What is the absolute deviation for 62 in the data set?
makvit [3.9K]
D is the answer for this question:))
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3 years ago
Guys Urgent Help me this example3.9 plzzz
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Answer:

A Hope this helps!!

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snow_lady [41]

Answer:

c. 16/35

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3 years ago
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The maximum afternoon temperature in Granderson is modeled by t=60-30 cos (x<img src="https://tex.z-dn.net/?f=%20%5Cpi%20" id="T
kicyunya [14]
The equation is t=60-30 cos (x \frac{\pi}{6})

the -30 expression, is subtracting, if negative, from the 60
if the cosine returned is negative, you'd end up with a +30,
negative * negative = positive

and then the 30 expression will ADD to the 60 amount
so the temperature "t" is highest, when the 30 expression, is
positive and and it's highest

when does that happen, when cosine is negative and at its highest,
well, cosine range is -1\le cos(\theta ) \le 1
so the lowest value cosine can provide is -1, when is cosine -1?
well, at \pi

so...  let's find a value that makes that expression to cos(\pi)

\bf t=60-30 cos (x \frac{\pi}{6}) \qquad x=6&#10;\\\\&#10;thus&#10;\\\\&#10;t=60-30 cos (6 \frac{\pi}{6})\implies t=60-30 cos (\pi )&#10;\\\\&#10;t=60-30[-1]\implies t=60+30\implies t=90

----------------------------------------------------------------------------------------
so...for August, that'll mean


\bf t=60-30 cos (x \frac{\pi}{6}) \qquad x=7&#10;\\\\&#10;t=60-30 cos (\frac{7\pi}{6})\implies t=60-30\left( -\cfrac{\sqrt{3}}{2} \right)&#10;\\\\&#10;t=60+(15\cdot \sqrt{3})\implies t\approx85.98^o


4 0
3 years ago
What is 2x-y=6 the slope
Flura [38]

First of all, lets get y on its own side of the equation.

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we can see that m=2, so the slope is 2 or \frac{2}{1}

5 0
3 years ago
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