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stiks02 [169]
3 years ago
11

Hows your day going:D?

Mathematics
2 answers:
kenny6666 [7]3 years ago
6 0

Answer:

yes.

Step-by-step explanation:

ollegr [7]3 years ago
6 0

Answer:

decent just doing hw wby

Step-by-step explanation:

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The FDA is testing a new ointment designed to deliver 3.5 micrograms of active ingredient to each square centimeter of skin. Nin
lions [1.4K]

Answer:

t(c)  =  3.355

Step-by-step explanation:

We assume a normal distribution, and with sample size n = 9 we should follow a t -student test on both tails since the FDA is interested in determining if the amount of drug absorbed is different from 3.5 micrograms.

Therefore if α  = 0,01    that means that confidence interval is 99 % or 0,99

Finally with   α/2 = 0,005   and 8 degrees of freedom we find in t-student table  t(c)  =  3.355

8 0
2 years ago
What is the same of the data? <br> skewed right<br> skewed left<br> symmetrical<br> not enough info
Solnce55 [7]
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2 years ago
Can someone please please help with these questions?
wolverine [178]
The answer to number seven is EDF!
5 0
3 years ago
PLEASE HELP ASAP!!!!!!!!!!!
irina [24]

Answer:

132 in^2 but DOUBLE CHECK MY WORK!!!!

Step-by-step explanation:

Break it down into parts ok. So I see a 10x10 square, two 2x4 right triangles and a 4x6 rectangle.

10x10 is 100 in

1/2(2*4) = 4 x 2 = 8 in

6x4 = 24 in

add them together to get 132 in^2

7 0
2 years ago
HELP PLEASE!! I DONT UNDERSTAND!!!!!!!!!! THANKS SO MUCH
8090 [49]

Hello, please consider the following.

We will multiply the numerator and denominator by

4+\sqrt{6x}

to get rid of the root in the denominator.

First of all, we cannot divide by 0, right? So, we need to make sure that the denominator is different from 0.

4-\sqrt{6x} =0\sqrt{6x}=4\\\\\text{Take the square}\\\\6x=4^2=16\\\\x=\dfrac{16}{6}=\dfrac{8}{3}

We need to take any x real number different from 8/3 then and simplify the expression.

Let's do it!

\begin{aligned}\dfrac{4}{4-\sqrt{6x}}&=\dfrac{4(4+\sqrt{6x})}{(4+\sqrt{6x})(4-\sqrt{6x})}\\\\&=\dfrac{4(4+\sqrt{6x})}{(4^2-\sqrt{6x}^2)}\\\\&=\dfrac{4(4+\sqrt{6x})}{(16-6x)}\\\\&=\dfrac{2(4+\sqrt{6x})}{(8-3x)}\\\\&\large \boxed{=\dfrac{8+2\sqrt{6x}}{8-3x}}\end{aligned}

Thank you

8 0
2 years ago
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