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Ostrovityanka [42]
2 years ago
14

Sindhu’s present age is thrice of Shilpa. If Shilpa’s age three years ago was x, then Sindhu’s present age is

Mathematics
2 answers:
Rudiy272 years ago
7 0

Answer:

I think "D"- 3(x + 3)

Step-by-step explanation:

as  3 years ago - x

so present age of shilpa  - x +3

~~~~~~~~~~~~~

present age  of sindhu - 3 (x+3)

i guess : )

Setler [38]2 years ago
5 0

Shilpa's age three years ago

Shilpa's age now = x + 3

Sindhu's age now = 3(x + 3)

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pentagon [3]

Answer:

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10-1/3

9/3

3=3

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The radius of a circle is 1 centimeter. What is the circle's circumference?
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6.28 cm

Step-by-step explanation:

The formula for the circumference of a circle with radius r is C = 2*pi*r.

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lapo4ka [179]

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Step-by-step explanation:

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"A study of the amount of time it takes a mechanic to rebuild the transmission for a 2005 Chevrolet Cavalier shows that the mean
Arada [10]

Answer:

85.31% probability that their mean rebuild time exceeds 8.1 hours.

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value 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. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central limit theorem:

The Central Limit Theorem estabilishes that, for a random variable X, with mean \mu and standard deviation \sigma, the sample means with size n of at least 30 can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}

In this problem, we have that:

\mu = 8.4, \sigma = 1.8, n = 40, s = \frac{1.8}{\sqrt{40}} = 0.2846

If 40 mechanics are randomly selected, find the probability that their mean rebuild time exceeds 8.1 hours.

This is 1 subtracted by the pvalue of Z when X = 8.1. So

Z = \frac{X - \mu}{\sigma}

By the Central Limit Theorem

Z = \frac{X - \mu}{s}

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Z = -1.05

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1 - 0.1469 = 0.8531

85.31% probability that their mean rebuild time exceeds 8.1 hours.

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