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labwork [276]
4 years ago
8

Please help me!!! Given that

Mathematics
1 answer:
Alinara [238K]4 years ago
3 0
Now we're talking. Imma go ahead and say that it's B
You might be interested in
Problem 2:
kifflom [539]

Answer:

The answers to the questions about problem 2 are:

  1. The fraction of the length after Priya stops two times is 3/4.
  2. The fraction of the length after Priya stops four times is 15/16.
  3. Priya will never reach the end of the hallway.

Step-by-step explanation:

The explanation about each answer is below:

1. In the first stop, Priya has walked half of the total length, and there would still be half the hallway, however, when she stops the second time, she has walked the half of the half, I mean:

\frac{1}{2}/2=\frac{1}{4}

If you add the two values you obtain the total distance traveled by Priya:

\frac{1}{2}+\frac{1}{4}= \frac{3}{4}

And the distance traveled in two stops is 3/4 of the total length of the hallway.

2. With four stops the system is the same, we know with two stops Priya travels 3/4 of the hallway, now with the third stop, she will travel half of the remaining distance:

\frac{1}{4}/2=\frac{1}{8}

And in the fourth stop she will travel half of the remaining, I mean:

\frac{1}{8}/2=\frac{1}{16}

Now, we add all the values of the distances obtained:

\frac{3}{4}+\frac{1}{8}+\frac{1}{16}= \frac{15}{16}

So, the distance traveled by Priya in the fourth stop is 15/16 of the total length of the hallway.

3. How we know, the numbers are infinite, in the same forms the distances, by this reason, how the problem says that Priya walks just half of the distance, she will never reach the end because despite she has very near of the end, she will continue walking just half and ever smaller distances.

3 0
3 years ago
Is 1/4 greater than 3/4
Eduardwww [97]
No because 3/4 is closer to one so 3/4 is greater
8 0
3 years ago
4(x-11)^2=25
netineya [11]
Expand the square: 4(x-11)^ =25
4(x-11)(x-11)=25

Distribute : 4(x-11)(x-11)=25
4(x(x-11)-11(x-11))=25

Distribute: 4(x(x-11)-11(x-11))=25
4(x^-11x-11(x-11))=25

Solution: x=17/2
x=27/2

hope that helps
8 0
3 years ago
Order the numbers in the group from least to greatest. 2.01, 2.1, 2.001
defon
The answer should be B.
4 0
3 years ago
Read 2 more answers
Orange M&M’s: The M&M’s web site says that 20% of milk chocolate M&M’s are orange. Let’s assume this is true and set
SOVA2 [1]

Answer:

The correct option is (A).

Step-by-step explanation:

Let <em>X</em> = number of orange  milk chocolate M&M’s.

The proportion of orange milk chocolate M&M’s is, <em>p</em> = 0.20.

The number of candies in a small bag of milk chocolate M&M’s is, <em>n</em> = 55.

The event of an milk chocolate M&M being orange is independent of the other candies.

The random variable <em>X</em> follows a Binomial distribution with parameter <em>n</em> = 55 and <em>p</em> = 0.20.

The expected value of a Binomial random variable is:

E(X)=np

Compute the expected number of orange  milk chocolate M&M’s in a bag of 55 candies as follows:

E(X)=np

         =55\times 0.20\\=11

It is provided that in a randomly selected bag of milk chocolate M&M's there were 14 orange ones, i.e. the proportion of orange milk chocolate M&M's in a random bag was 25.5%.

This proportion is not surprising.

This is because the average number of orange milk chocolate M&M’s in a bag of 55 candies is expected to be 11. So, if a bag has 14 orange milk chocolate M&M’s it is not unusual at all.

All unusual events have a very low probability, i.e. less than 0.05.

Compute the probability of P (X ≥ 14) as follows:

P(X\geq 14)=\sum\limits^{55}_{x=14}{{55\choose x}0.20^{x}(1-0.20)^{55-x}}

                 =0.1968

The probability of having 14 or more orange candies in a bag of milk chocolate M&M’s is 0.1968.

This probability is quite larger than 0.05.

Thus, the correct option is (A).

4 0
4 years ago
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