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2 years ago

8

PLEASE HELP ME QUICKKK, FIRST CORRECT PERSON GETS BRAINLIEST

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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There are seven cats in my neighborhood, with an average of $41$ whiskers each. If one of the cats has $17$ whiskers, how many w

harkovskaia [24]

Answer:

Each of the other cats have 45 whiskers on average.

Step-by-step explanation:

Let x represent the average number of whisker of each cat.

We have been given that there are 7 cats in my neighborhood, with an average of 41 whiskers each.

The total number of whiskers of six cats would be $6x$ .

Since one of the cats has 17 whiskers, so the total number of whiskers of 7 cats would be $6x+17$

We will use average formula to solve our given problem.

$\text{Average}=\frac{\text{Sum of data points}}{\text{Number of data points}}$

Upon substituting our given values, we will get:

$41=\frac{6x+17}{7}$

Let us solve for x.

$41*7=\frac{6x+17}{7}*7$

$287=6x+17$

$287-17=6x+17-17$

$270=6x$

Switch sides:

$6x=270$

$\frac{6x}{6}=\frac{270}{6}$

$x=45$

Therefore, the each of the other cats have 45 whiskers on average.

8 0

3 years ago

Twice a number is increased by 3 is 7

lys-0071 [83]

7-3=4

4/2=2

The answer is 2

6 0

3 years ago

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Is this a function? <br> A) yes<br> B) No

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Don’t listen to get the links they are annoying

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2 years ago

How to do a do and undo chart for 3s=k(b+c)

katen-ka-za [31]

Answer:5

Step-by-step explanation:

4 0

3 years ago

How derivative of this absolute value function is like this!

Daniel [21]

Recall the definition of absolute value.

$|x| = \begin{cases} x & \text{if } x \ge 0 \\ -x & \text{if } x < 0 \end{cases}$

When $x$ ,

$|x| = x \implies \dfrac{d|x|}{dx} = 1$

When $x$ ,

$|x| = -x \implies \dfrac{d|x|}{dx} = -1$

The derivative does not exist at $x=0$ , since the one-sided limits

$\displaystyle \lim_{x\to0^-} f'(x) = -1$

and

$\displaystyle \lim_{x\to0^+} f'(x) = +1$

do not match.

So the derivative of $|x|$ is

$\dfrac{d|x|}{dx} = \begin{cases} 1 & \text{if } x > 0 \\ -1 & \text{if } x < 0 \\ \text{undefined} & \text{if } x = 0\end{cases}$

Now we can write this as

$\dfrac{d|x|}{dx} = \dfrac x{|x|} = \dfrac{|x|}x$

since

$x > 0 \implies |x| = x \implies \dfrac{|x|}x = \dfrac xx = 1$

and

$x < 0 \implies |x| = -x \implies \dfrac{|x|}x = -\dfrac xx = -1$

6 0

1 year ago