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viktelen [127]
2 years ago
5

Rewrite using a single exponent

Mathematics
1 answer:
Vika [28.1K]2 years ago
7 0

Can you send a picture? what is the question?

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How many groups of 12 are in 600
Ugo [173]
Do 600÷12= 50

The answer is 50 
4 0
3 years ago
Read 2 more answers
Find the arc length with the given information.<br> central angle = n/4, radius=3
BabaBlast [244]

Answer:

\frac{3\pi }{4}

Step-by-step explanation:

Arc length is the product of central angle(in radians) and radius.

i.e.

arc length=radius *central angle(in radians)

s=r*theta\\s=3*\frac{\pi }{4}\\ s=\frac{3\pi }{4}

7 0
3 years ago
Read 2 more answers
Solve.<br> −8.5d ≥ 20<br><br> d ≤ –170<br> d ≥ –170
saw5 [17]

Answer:

\rm =>D \leqslant 2.352941

Step-by-step explanation:

−8.5d ≥ 20

Divide both sides by -8.5:

=> \cfrac {- 8.5 \: d}{  - 8.5} \ \geqslant  \cfrac{20}{ - 8.5}

\rm=>d \leqslant 2.352941

This is the answer.

3 0
2 years ago
What is the answer to 3(2-t)/4 = 3t/4
egoroff_w [7]

Answer:

t=1

Step-by-step explanation:

First we distribute

[3(2-t)]/4=3t/4

(6-3t)/4=3t/4

Then we multiply both sides by 4 to cancel out the denominator

6-3t=3t

Add 3t to both sides

6t=6

Divide both sides by 6

t=1

8 0
3 years ago
From a sample with n=32, the mean duration of a geyser's eruption is 3.81 minutes and the standard deviation is 1.07 minutes. Us
Damm [24]
Notice that

1.67=3.81-2\times1.07
5.95=3.81+2\times1.07

This means you are looking for

\mathbb P(1.67

Chebyshev's theorem states that for any constant k>0,

\mathbb P(|X-\mu|

which means

\mathbb P(1.67

In other words, at least 75% of the geysers will have eruptions lasting between 1.67 and 5.95 minutes. Out of 32 geysers, then, one can expect at least 24 of them to have eruptions with between these durations.
5 0
3 years ago
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