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zloy xaker [14]
3 years ago
10

What is he interquartile range for 27,4,54,78,27,48,79,64,5,6,41,71

Mathematics
2 answers:
Ugo [173]3 years ago
7 0

Answer:

If it stays in that order its between 48,79 if this helps.

Step-by-step explanation:

lorasvet [3.4K]3 years ago
4 0

Ok if you have the K12 quiz the answer this question is 51.

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For what real values of does the quadratic 12x^2+kx+27=0 have nonreal roots
bija089 [108]

Answer:

k<±36

Step-by-step explanation:

Δ<0 (no real roots)

b²-4ac<0

k²-4x12x27<0

k²-1296<0

k²<1296

k<±36

3 0
3 years ago
PLEASE HELP ASAP!!! i dont understand :(
aivan3 [116]

Answer:

hello,

Step-by-step explanation:

y=k(x-a')²+b'

focus is (a',1/(4k)+b')

2x²=-120y

==> y=-2/120x²

==> y=-1/60(x-0)²+0

focus= (0,-15)

7 0
3 years ago
(3.9 × 10 8)(8.2 × 10‐2) =? ANSWEERRR PLSS
Scorpion4ik [409]

Answer:

its 2480

its 2,480

Step-by-step explanation:

thatss my answerrr

7 0
3 years ago
Here we will study the function f (x) = e ^ x sin (x), where x ∈ [0, 2π]. a) Determine where the function is decreasing and incr
Semmy [17]

fff

f(x) = e^x sin (x)

To find increasing and decreasing intervals we take derivative

f'(x) = e^xsin(x)+e^x(cosx)= e^x(sinx+cosx)

Now we set the derivative =0  and solve for x

e^x(sinx+cosx)=0

sinx + cosx =0

divide whole equation by cos x

\frac{sinx}{cosx} + \frac{cosx}{cosx} =0

tanx +1 =0

tanx = 1

x=\frac{3\pi }{4} and  x=\frac{7\pi}{4}

Now we pick a number between 0 to  \frac{3\pi }{4}

Lets pick  \frac{\pi }{2}

Plug it into the derivative

f'(x) =e^{\frac{\pi }{2}}(sin(\frac{\pi}{2})+cos(\frac{\pi }{2}))

= 4.810 is positive

So the graph of f(x) is increasing on the interval [0, x=\frac{3\pi }{4})

Now we pick a number between   \frac{7\pi}{4} to 2pi

Lets pick  \frac{11\pi}{6}

Plug it into the derivative

f'(x) =e^{\frac{11\pi}{6}}(sin(\frac{11\pi}{6})+cos(\frac{11\pi }{6}))

= 116 is positive

So the graph of f(x) is increasing on the interval (\frac{7\pi }{4}, 2\pi)

Increasing interval is (0,\frac{3\pi }{4}) U (\frac{7\pi }{4}, 2\pi)

Decreasing interval is (\frac{3\pi}{4}, \frac{7\pi}{4})

(b)

The graph of f(x) increases and reaches a local maximum at x=\frac{3\pi}{4}

The graph of f(x) decreases and reaches a local minimum at x=\frac{7\pi}{4}

(c)

f(0) = 0

f(2\pi)=0

f(\frac{3\pi }{4})=7.46

f(\frac{7\pi}{4})=-172.64

Here global maximum at x=\frac{3\pi}{4}

Here global minimum at x=\frac{7\pi}{4}


3 0
3 years ago
Read 2 more answers
Help ASAP quick answer please!!!
Volgvan
The answer is the second option, -0.5
8 0
4 years ago
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