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ArbitrLikvidat [17]
3 years ago
6

Please answer quickly

Mathematics
1 answer:
storchak [24]3 years ago
5 0

Answer:

1 1/2

Step-by-step explanation:

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The response to a question has three alternatives: A, B, and C. A sample of 120 responses provides 60 A, 12 B, and 48 C. Show th
zimovet [89]

Answer:

CLASS     FREQUENCIES     RELATIVE FREQUENCIES

A                        60                                 0.5

B                        12                                  0.1

C                        48                                 0.4

TOTAL              120                                  1

Step-by-step explanation:

Given that;

the frequencies of there alternatives are;

Frequency A = 60

Frequency B = 12

Frequency C = 48

Total = 60 + 12 + 48 = 120

Now to determine our relative frequency, we divide each frequency by the total sum of the given frequencies;

Relative Frequency A = Frequency A / total = 60 / 120 = 0.5

Relative Frequency B = Frequency B / total = 12 / 120 = 0.1

Relative Frequency C = Frequency C / total = 48 / 120 = 0.4

therefore;

CLASS     FREQUENCIES     RELATIVE FREQUENCIES

A                        60                                 0.5

B                        12                                  0.1

C                        48                                 0.4

TOTAL              120                                  1

5 0
3 years ago
Problem PageQuestion The mass of a radioactive substance follows a continuous exponential decay model, with a decay rate paramet
asambeis [7]

Answer:

<em>t = 1.51</em>

Step-by-step explanation:

<u>Exponential Model</u>

The exponential model is often used to simulate the behavior of a magnitude that either grow or decay in proportion to the existing amount of that magnitude.

The model can be expressed as

M=M_oe^{kt}

In this case, Mo is the initial mass of the radioactive substance and k is a constant which value is positive if the mass is growing or negative if the mass is decaying.

The value of k is not precisely given in the question, we are assuming k=-0.2

The model is now

M=M_oe^{-0.2t}

We are required to compute the time it takes the mass to reach one-half of its initial value:

\displaystyle \frac{M_o}{2}=M_oe^{-0.2t}

Simplifying

\displaystyle \frac{1}{2}=e^{-0.2t}

Taking logarithms

\displaystyle ln\frac{1}{2}=ln(e^{-0.2t})=-0.2t

Solving for t

\displaystyle t=-\frac{ln\frac{1}{2}}{0.2}=1.51

6 0
3 years ago
Ooooooooooooooooooooooooooo
Klio2033 [76]

Answer:

yes

Step-by-step explanation:

8 0
3 years ago
Read 2 more answers
Complement and supplement of a 19
Tems11 [23]

well first the complement I 90-19=71

and supplement is 180-19=161

3 0
3 years ago
Any answers? No need for an explanation if you want to sure, the answer would be appreciated!
Simora [160]
Yes because 9+4 = 13 and if you plug in the y and x it turns out true
6 0
3 years ago
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