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3 years ago

What does e=mc3 mean?

1 answer:

5 0

It is an equation of energy and mass which was given by Albert Einstein
and it is e=mc2 which means that energy and mass are interchangable
where
 E is energy.
M is mass.
C is the speed of light
hope it helps

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Is 1/3, 2/3, 4/3, 8/3, 16/3 arithmetic sequence

11111nata11111 [884]

The given sequence is not arithmetic sequence

Solution:

Given sequence is:

$\frac{1}{3} , \frac{2}{3} , \frac{4}{3} , \frac{8}{3} , \frac{16}{3}$

We have to find if the above sequence is arithmetic sequence or not

An arithmetic sequence is a sequence of numbers such that the difference of any two successive members of the sequence is a constant

Here in the given sequence

$\text{ first term } = a_1 = \frac{1}{3}\\\\\text{ second term } = a_2 = \frac{2}{3}\\\\\text{ third term } = a_3 = \frac{4}{3}\\\\\text{ fourth term } = a_4 = \frac{8}{3}\\\\\text{ fifth term } = a_5 = \frac{16}{3}$

Let us find the difference between terms

$a_2 - a_1 = \frac{2}{3} - \frac{1}{3} = \frac{1}{3}$

$a_3 - a_2 = \frac{4}{3} - \frac{2}{3} = \frac{2}{3}$

$a_4 - a_3 = \frac{8}{3} - \frac{4}{3} = \frac{4}{3}$

$a_5 - a_4 = \frac{16}{3} - \frac{8}{3} = \frac{8}{3}$

Thus the difference between terms is not constant

So the given sequence is not arithmetic sequence

5 0

3 years ago

The route used by a certain motorist in commuting to work contains two intersections with traffic signals. The probability that

tekilochka [14]

Answer:

a) P(X∩Y) = 0.2

b) $P_1$ = 0.16

c) P = 0.47

Step-by-step explanation:

Let's call X the event that the motorist must stop at the first signal and Y the event that the motorist must stop at the second signal.

So, P(X) = 0.36, P(Y) = 0.51 and P(X∪Y) = 0.67

Then, the probability P(X∩Y) that the motorist must stop at both signal can be calculated as:

P(X∩Y) = P(X) + P(Y) - P(X∪Y)

P(X∩Y) = 0.36 + 0.51 - 0.67

P(X∩Y) = 0.2

On the other hand, the probability $P_1$ that he must stop at the first signal but not at the second one can be calculated as:

$P_1$ = P(X) - P(X∩Y)

$P_1$ = 0.36 - 0.2 = 0.16

At the same way, the probability $P_2$ that he must stop at the second signal but not at the first one can be calculated as:

$P_2$ = P(Y) - P(X∩Y)

$P_2$ = 0.51 - 0.2 = 0.31

So, the probability that he must stop at exactly one signal is:

$P = P_1+P_2\\P=0.16+0.31\\P=0.47$

7 0

2 years ago

Compare the values of each 3 in the number 633,248

ipn [44]

633,248.
The difference of each 3 is that they have different place values.
The 3 on the right side is int he 1000s (thousands) place, and the 3 on the left is in the 10,000s (ten thousands) place. :)

The one in the 10,000s place is 10 times larger than the one in the 1,000s place!

6 0

2 years ago

Complete the pattern 274÷1=

Orlov [11]

The answer 274 because anything divided by one is itself

5 0

2 years ago

Please help me I will give brain and I will be so happy btw 20 points

iVinArrow [24]

Answer:

The first answer and the second answer.

5 0

2 years ago

Read 2 more answers