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3 years ago

If the decimal is 0.121122111222... if the pattern keeps going will it be a repeating decimal??

Mathematics

2 answers:

sveta [45]3 years ago

7 0

No. A repeating decimal is when it is a repeated number. For example, 0.633333333333 is a repeated decimal.

Doss [256]3 years ago

5 0

First we observe pattern here. The pattern is one number "1" and one number "2" than two numbers "1" and two numbers "2" and so on. Every next array of numbers 1 and 2 will have one extra member.
Because of this, 0.121122111222... isn't repeating decimal. Repeating decimal has some number of decimals that keep repeating. For example if we had:0.121122 121122 121122 ... this would be repeating decimal because 121122 segment keeps repeating.
In your task you can't find a segment that keeps repeating.
hope this helps

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Answer:

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Artemon [7]

Answer:

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Step-by-step explanation:

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Which of the following is a correct tangent ratio for the figure?

Ludmilka [50]

Answer:

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3 years ago

A giraffe can travel 800 feet in 20 seconds. Write a direct variation equation for the distance traveled any time | Please expla

Diano4ka-milaya [45]

The direct variation equation is:
d=kt
where d is the distance that the giraffe travels
k is the constant of variation
and t is time
Now we are told that the giraffe can travel 800 ft in 20 seconds, so we can solve for the constant of variation (k)
800=k(20) divide both sides by 20
k=40-------------and the units are ft per sec
So now we can write
d=40t
ck
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3 years ago

The service department of a luxury car dealership conducted research on the amount of time its service technicians spend on each

mart [117]

Answer:

Probability that the mean service time is between 1 and 2 hours is 0.96764.

Step-by-step explanation:

We are given that a systematic random sample of 100 service appointments has been collected.

The 100 appointments showed an average preparation time of 90 minutes with a standard deviation of 140 minutes.

Let $\bar X$ = sample mean service time

The z-score probability distribution for sample mean is given by;

Z = $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, $\mu$ = average preparation time = 90 minutes

$\sigma$ = standard deviation = 140 minutes

n = sample of appointments = 100

The Z-score measures how many standard deviations the measure is away from the mean. After finding the Z-score, we look at the z-score table and find the p-value (area) associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X.

Now, probability that the mean service time is between 60 and 120 minutes is given by = P(60 minutes < $\bar X$ < 120 minutes)

P(60 minutes < $\bar X$ < 120 minutes) = P( $\bar X$ < 120 min) - P( $\bar X$ $\leq$ 60 min)

P( $\bar X$ < 120 min) = P( $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ < $\frac{120-90}{\frac{140}{\sqrt{100} } }$ ) = P(Z < 2.14) = 0.98382

P( $\bar X$ $\leq$ 60 min) = P( $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ $\leq$ $\frac{60-90}{\frac{140}{\sqrt{100} } }$ ) = P(Z $\leq$ -2.14) = 1 - P(Z < 2.14)

= 1 - 0.98382 = 0.01618

The above probability is calculated by looking at the value of x = 2.14 in the z table which has an area of 0.98382.

Therefore, P(60 min < $\bar X$ < 120 min) = 0.98382 - 0.01618 = 0.96764

7 0

3 years ago