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marysya [2.9K]
4 years ago
5

An adolescent has a weight of 127 pounds. The doctor requires you to record the patient’s weight in kilograms. Convert 127 pound

s to kilograms.
Mathematics
2 answers:
olasank [31]4 years ago
6 0

The amount of kilograms he weighs I think is 57.6062994983

Soloha48 [4]4 years ago
3 0

Answer:

57.606184 kilograms

Step-by-step explanation:

Given :An adolescent has a weight of 127 pounds.

To Find : Convert 127 pounds to kilograms.

Solution :

Weight of adolescent = 127 pounds

We are supposed to find the weight in kilograms.

1 pound = 0.453592 kilograms

127 pounds = 0.453592 \times 127 kilograms

127 pounds = 57.606184 kilograms

Hence 127 pounds in kilograms is 57.606184 kilograms

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Read 2 more answers
P(X< ) 1-P(X> ) A softball pitcher has a 0.626 probability of throwing a strike for each curve ball pitch. If the softball
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Answer:

19.49% probability that no more than 16 of them are strikes

Step-by-step explanation:

Binomial probability distribution

Probability of exactly x sucesses on n repeated trials, with p probability.

Can be approximated to a normal distribution, using the expected value and the standard deviation.

The expected value of the binomial distribution is:

E(X) = np

The standard deviation of the binomial distribution is:

\sqrt{V(X)} = \sqrt{np(1-p)}

Normal probability distribution

Problems of normally distributed samples can be 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.

When we are approximating a binomial distribution to a normal one, we have that \mu = E(X), \sigma = \sqrt{V(X)}.

In this problem, we have that:

n = 30, p = 0.626

So

\mu = E(X) = np = 30*0.626 = 18.78

\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{30*0.626*0.374} = 2.65

What is the probability that no more than 16 of them are strikes?

Using continuity correction, this is P(X \leq 16 + 0.5) = P(X \leq 16.5), which is the pvalue of Z when X = 16.5. So

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

Z = \frac{16.5 - 18.78}{2.65}

Z = -0.86

Z = -0.86 has a pvalue of 0.1949

19.49% probability that no more than 16 of them are strikes

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