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shutvik [7]
3 years ago
11

Wally had two more marbled than ashan and together they have 88 marbles how many marbles does ashan have

Mathematics
2 answers:
labwork [276]3 years ago
5 0
Represent as
w=wally
a=ashan
so wally has 2 more marbles than ashan
w=2+a

together they have 88
w+a=88

since w=2+a, subsitute (2+a) for w in the w+a=88 equation
2+a+a=88
add like terms
2+2a=88
subtract 2 from both sides
2a=86
divide by 2
a=43

ashan has 43 marbles
Naily [24]3 years ago
4 0
The obvious answer would be 86, but you have to look closely at the problem. If Wally had two more marbles than Ashan, then that would mean Wally has 45. If Ashan has 86 then that is incorrect because Wally only has two, not two more than Ashan. If Wally has 45, then 80-45 is 43. This equation now makes sense because Wally has two more marbles than ashen, so Wally has 45 marbles and Ashan has 43.
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For a certain river, suppose the drought length Y is the number of consecutive time intervals in which the water supply remains
AnnZ [28]

Answer:

a) There is a 9% probability that a drought lasts exactly 3 intervals.

There is an 85.5% probability that a drought lasts at most 3 intervals.

b)There is a 14.5% probability that the length of a drought exceeds its mean value by at least one standard deviation

Step-by-step explanation:

The geometric distribution is the number of failures expected before you get a success in a series of Bernoulli trials.

It has the following probability density formula:

f(x) = (1-p)^{x}p

In which p is the probability of a success.

The mean of the geometric distribution is given by the following formula:

\mu = \frac{1-p}{p}

The standard deviation of the geometric distribution is given by the following formula:

\sigma = \sqrt{\frac{1-p}{p^{2}}

In this problem, we have that:

p = 0.383

So

\mu = \frac{1-p}{p} = \frac{1-0.383}{0.383} = 1.61

\sigma = \sqrt{\frac{1-p}{p^{2}}} = \sqrt{\frac{1-0.383}{(0.383)^{2}}} = 2.05

(a) What is the probability that a drought lasts exactly 3 intervals?

This is f(3)

f(x) = (1-p)^{x}p

f(3) = (1-0.383)^{3}*(0.383)

f(3) = 0.09

There is a 9% probability that a drought lasts exactly 3 intervals.

At most 3 intervals?

This is P = f(0) + f(1) + f(2) + f(3)

f(x) = (1-p)^{x}p

f(0) = (1-0.383)^{0}*(0.383) = 0.383

f(1) = (1-0.383)^{1}*(0.383) = 0.236

f(2) = (1-0.383)^{2}*(0.383) = 0.146

Previously in this exercise, we found that f(3) = 0.09

So

P = f(0) + f(1) + f(2) + f(3) = 0.383 + 0.236 + 0.146 + 0.09 = 0.855

There is an 85.5% probability that a drought lasts at most 3 intervals.

(b) What is the probability that the length of a drought exceeds its mean value by at least one standard deviation?

This is P(X \geq \mu+\sigma) = P(X \geq 1.61 + 2.05) = P(X \geq 3.66) = P(X \geq 4).

We are working with discrete data, so 3.66 is rounded up to 4.

Either a drought lasts at least four months, or it lasts at most thee. In a), we found that the probability that it lasts at most 3 months is 0.855. The sum of these probabilities is decimal 1. So:

P(X \leq 3) + P(X \geq 4) = 1

0.855 + P(X \geq 4) = 1

P(X \geq 4) = 0.145

There is a 14.5% probability that the length of a drought exceeds its mean value by at least one standard deviation

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James Wrote the number 8,980,000 in scientific notation which number did he write
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<span>8,980,000 = 8.98 x 10</span>⁶
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8 0
2 years ago
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What is the midpoint of a segment with endpoints at(-4,-8) and (8, 10)?
vekshin1

Answer:

The answer is

<h2>( 2 , 1)</h2>

Step-by-step explanation:

The midpoint M of two endpoints of a line segment can be found by using the formula

M = ( \frac{x1 + x2}{2}  , \:  \frac{y1 + y2}{2} )

where

(x1 , y1) and (x2 , y2) are the points

From the question the points are

(-4,-8) and (8, 10)

The midpoint is

M = ( \frac{ - 4 + 8}{2}  , \:  \frac{ - 8 + 10}{2} ) \\  = ( \frac{4}{2} ,  \frac{2}{2} )

We have the final answer as

<h3>( 2 , 1)</h3>

Hope this helps you

4 0
3 years ago
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