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NNADVOKAT [17]
3 years ago
11

I NEED HELP PLEASE I GIVE 5 STARS !

Mathematics
2 answers:
uranmaximum [27]3 years ago
8 0

Answer:

C. 2\sqrt{29}

Step-by-step explanation:

Square root of 116 is 10.7703296

Square root of 29 is 5.38516481, but as it is multiplied by 2, it becomes 10.7703296

vagabundo [1.1K]3 years ago
7 0

Answer:

C

Step-by-step explanation:

The square root of 116 is about ten. Two times the square root of 29 also equals about ten. A equals about 13. B equals about 9. And 13 equals 13 so only C can be correct.

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If 4m =130 does 4m÷4=130
Simora [160]
No because you would be dividing the first equation by four, for an example if m=32.5 in the first equation, if you put that into the second equation and divide it by four you won’t get 130 like in the first equation
5 0
3 years ago
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Solve using logs 5^(3x-6)=125
Hitman42 [59]
5 ^{3x-6} =125 \\ \\ 5 ^{3x-6} =5^3 \\ \\3x-6 =3 \ \ |+6 \\ \\3x=3+9 \\ \\3x=12 \ \ /:3 \\ \\x=\frac{12}{3} \\ \\x= 3
3 0
3 years ago
I need help plizzzzz
Nataly [62]
Try this dude i hope it is right
5 0
3 years ago
Find the probability that the person is frequently or occasionally involved in charity work.
Schach [20]
Given the table below which shows the result of a survey that asked 2,881 people whether they are involved in any type of charity work.

\begin{tabular}
{|c|c|c|c|c|c|}
 &Frequently&Occassionally&Not at all&Total\\[1ex]
Male&227&454&798&1,479\\
Female &205&450&747&1,402\\
Total&432&904&1,545&2,881
\end{tabular}

Part A:

If a person is chosen at random, the probability that the person is frequently or occassinally involved in charity work is given by

P(being \ frequently \ involved \ or \ being \ occassionally \ involved)\\ \\= \frac{432}{2881} + \frac{904}{2881} = \frac{1336}{2881}=\bold{0.464}



Part B:

If a person is chosen at random, the probability that the person is female or not involved in charity work at all is given by

P(being
 \ female \ or \ not \ being \ involved)\\ \\= 
\frac{1402}{2881} + \frac{1545}{2881}-\frac{747}{2881} = 
\frac{2200}{2881}=\bold{0.764}



Part C:

If a person is chosen at random, the probability that the person is male or frequently involved in charity work is given by

P(being
 \ male \ or \ being \ frequently \ involved)\\ \\= 
\frac{1479}{2881} + \frac{432}{2881}-\frac{227}{2881} = 
\frac{1684}{2881}=\bold{0.585}



Part D:

If a person is chosen at random, the probability that the person is female or not frequently involved in charity work is given by

P(being
 \ female \ or \ not \ being \ frequently \ involved)\\ \\= 
\frac{1402}{2881} + \frac{904}{2881} + \frac{1545}{2881}-\frac{450}{2881}-\frac{747}{2881} = 
\frac{2654}{2881}=\bold{0.921}



Part E:

The events "being female" and "being frequently involved in charity work" are not mutually exclusive because being a female does not prevent a person from being frequently involved in charity work.

Indeed from the table, there are 205 females who are frequently involved in charity work.

Therefore, the answer to the question is "No, because 205 females are frequently involved charity work".
4 0
3 years ago
Plssssssss help solve
Lana71 [14]

Answer:

cosine = adjacent/hypotenuse

cos A = 20/29   (choice: yellow)

Step-by-step explanation:

3 0
3 years ago
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