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s344n2d4d5 [400]

2 years ago

7

geo drinks 36 ounces of water per day. he reads that he should drink 64 ounces of water per day. if he starts drinking 64 ounces

, what is the percent increase? round to the nearest percent.

1 answer:

andreyandreev [35.5K]2 years ago

3 0

Your answer is 23%, hope this helps

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A) 2.5x + 1.25 < 3.75 <br>b)5.12 2.4x + 0.3<br>help!!!<br><br>

Sauron [17]

Answer:

a) 1

b)I would like to help u but i don't understand the question I think something is missing there plus sign or minus sign sry

Step-by-step explanation:

2.5x + 1.25 < 3.75

2.5x < 3.75 - 1.25

2.5x < 2.5

$\frac{2.5x}{2.5} = \frac{2.5}{2.5}$

x = 1

I hope u like it ‍❤️‍‍❤️‍

For any question comment me

Really sorry for the second question

4 0

3 years ago

A small point deduction applies if a participation activity's question is not answered correctly the first time?

LiRa [457]

Answer: False

Step-by-step explanation:

There is no such deduction when a participation's questions are not answered correctly the first time. Whatever answer is given is part of the learning curve and ensures that the activity can be improved upon.

Had there been a small point deduction then there would be no opportunity to learn because there would be too much fear associated with the wrong answer.

3 0

2 years ago

2 2/9+ 3 3/7 +1 4/9 enter sum in a box as a mixed number in simplest form

viva [34]

Answer:

Input Data :

Mixed Fraction A =

7

1

6

Mixed Fraction B =

2

1

3

Step-by-step explanation:

wrong

4 0

3 years ago

Read 2 more answers

A coin, having probability p of landing heads, is continually flipped until at least one head and one tail have been flipped. (a

Natali [406]

Answer:

(a)

The probability that you stop at the fifth flip would be

$p^4 (1-p) + (1-p)^4 p$

(b)

The expected numbers of flips needed would be

$\sum\limits_{n=1}^{\infty} n p(1-p)^{n-1} = 1/p$

Therefore, suppose that $p = 0.5$ , then the expected number of flips needed would be 1/0.5 = 2.

Step-by-step explanation:

(a)

Case 1

Imagine that you throw your coin and you get only heads, then you would stop when you get the first tail. So the probability that you stop at the fifth flip would be

$p^4 (1-p)$

Case 2

Imagine that you throw your coin and you get only tails, then you would stop when you get the first head. So the probability that you stop at the fifth flip would be

$(1-p)^4p$

Therefore the probability that you stop at the fifth flip would be

$p^4 (1-p) + (1-p)^4 p$

(b)

The expected numbers of flips needed would be

$\sum\limits_{n=1}^{\infty} n p(1-p)^{n-1} = 1/p$

Therefore, suppose that $p = 0.5$ , then the expected number of flips needed would be 1/0.5 = 2.

7 0

2 years ago

Select the correct answer.

Irina18 [472]

D. About 122 units I just took the test

4 0

2 years ago