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kari74 [83]
3 years ago
8

PLEASE HELP ME NO LINKS PLEASE

Mathematics
2 answers:
Rufina [12.5K]3 years ago
5 0

Answer:

Step-by-step explanation:Class 1 :

mean= 78.83(rounded)  median=88 range= 55

Class 2:

mean= 79.42 (rounded) median=85.5 range= 60

Class 3:

mean= 79.42 (rounded) median=78 range =45

marissa [1.9K]3 years ago
4 0

Answer: Don't know sorry what kind of grade is this for

Step-by-step explanation:

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Connor and Angie helped take attendance during their schools practice fire drill if the actual count was between 77 and 89 inclu
wel

Answer:

The most absolute error is 12

Step-by-step explanation:

The formula for absolute error is given by the following relation;

Absolute error, Δx = Value measured, x₀ - Exact (actual) value, x

Δx = x₀ - x

Since the actual count is between 77 and 89, the maximum absolute error will be where the actual value = 77 and the measured value = 89 and vice versa, giving an absolute error of 89 -77 = 12.

The most absolute error = 12.

4 0
3 years ago
Find the surface area of the prism.
Katena32 [7]

Answer:

total surface area = 324 m²

lateral surface are = 132 m²

top surface area = 96 m²

bottom surface area = 96 m²

4 0
2 years ago
Which list has three equivalent fractions for 2/5
Aleksandr [31]

Answer:

The correct answer is B.

We have 4 / 10 = ( 2 x 2 ) / ( 2 x 5 );

6 / 15 = ( 3 x 2 ) / ( 3 x 5 );

8 / 20 = ( 4 x 2 ) / ( 4 x 5 );

Step-by-step explanation:

8 0
3 years ago
Read 2 more answers
Find the absolute maximum and absolute minimum values of f on the given interval. f(t) = 2 cos(t) + sin(2t),[0, π/2].
castortr0y [4]

Answer:

The absolute maximum is \frac{3\sqrt 3}2 and the absolute minimum value is 0.

Step-by-step explanation:

Differentiate of f both sides w.r.t.  t,

f(t)=2 \cos t+\sin 2t

\Rightarrow f'(t)=-2\sin t+2\cos 2t

Now take f'(t)=0

\Rightarrow -2\sin t+2\cos 2t=0

\Rightarrow 2\cos 2t=2\sin t

\Rightarrow \cos 2t=\sin t

\Rightarrow 1-2\sin ^2t =\sin t  \quad \quad  [\because \cos 2t = 1-2\sin ^2t]

\Rightarrow 2\sin ^2t+\sin t-1=0

\Rightarrow 2\sin ^2t+2\sin t-\sin t-1=0

\Rightarrow 2\sin t(\sin t+1)-1(\sin t+1)=0

\Rightarrow (\sin t+1)(2\sin t-1)=0

\Rightarrow \sin t+1=0  \;\text{and}\; 2\sin t-1=0

\Rightarrow \sin t =-1  \;\text{and}\;   \sin t =\frac 12

In the interval 0\leq t\leq \frac {\pi}2, the answer to this problem is \frac {\pi}6

Now find the second derivative of f(t) w.r.t.   t,

f''(t)=-2\cos t-4\sin 2t

\Rightarrow \left[f''(t)\right]_{t=\frac {\pi}6}=-2\times \frac {\sqrt 3}2-4\times \frac{\sqrt 3}2=-3\sqrt 3

Thus, f(t) is maximum at t=\frac {\pi}6 and minimum at t=0

\left[f(t)\right]_{t=\frac {\pi}6}=2\times \frac {\sqrt 3}2+\frac{\sqrt 3}2=\frac{3\sqrt 3}2\;\text{and}\;\left[f(t)\right]_{t=\frac{\pi}2}= 2\times 0+0=0

Hence, the absolute maximum is \frac{3\sqrt 3}2 and the absolute minimum value is 0.

7 0
3 years ago
Find the area. Answer ASAP pls pls
Aleonysh [2.5K]

Answer:

This is your answer ☺️

4 0
3 years ago
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