What is 6.46 6.11 and 4.6 in order from least to greatest

Sixty percent of those who attended the soccer game were parents. There were 120 people who attended the soccer game. How many p

iren [92.7K]

Answer: 48, 25%, 125 , 60, 80

Step-by-step explanation:

7 0

3 years ago

A card is Chosen at random from a deck of 52 cards it is replaced and a second card is chosen what is the probability both cards

ankoles [38]

There are 4 aces in a deck of 52 cards. the odds of drawing one is 1/13. the probability of doing it again is still 1/13. the odds of doing it twice, as described, is then 1/13 * 1/13, or (1/13)², or 1/169 option A

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3 years ago

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Line MN passes through points M(4, 3) and N(7, 12). If the equation of the line is written in slope-intercept form, y = mx + b,

expeople1 [14]

For finding the value of b, we must consider that Line MN passes through points M(4, 3) and N(7, 12). With this condition y = mx + b, can be written 3=4m+ b (because line passes through M(4,3) ) and 12=7m+b, b ( because line passes through M(7,12)). We have a system of equation 4m+ b=3 7m+b=12 For solving this, 4m+b- (7m+b)= 3-12, it is equivalent to -3m= -9 and then m=3, if m=3 so 4x3 +b =3 implies b= 3 -12= -9, so the value of b= -9 <span>
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3 0

3 years ago

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Write 47 and 1000 as a decimal

Marrrta [24]

47 is equal to .47 and 1000 is equal to 1

have a great day ♡

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3 years ago

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Suppose the horses in a large stable have a mean weight of 975lbs, and a standard deviation of 52lbs. What is the probability th

Lubov Fominskaja [6]

Answer:

0.8926 = 89.26% probability that the mean weight of the sample of horses would differ from the population mean by less than 15lbs if 31 horses are sampled at random from the stable

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 normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

In this problem, we have that:

$\mu = 975, \sigma = 52, n = 31, s = \frac{52}{\sqrt{31}} = 9.34$

What is the probability that the mean weight of the sample of horses would differ from the population mean by less than 15lbs if 31 horses are sampled at random from the stable?

pvalue of Z when X = 975 + 15 = 990 subtracted by the pvalue of Z when X = 975 - 15 = 960. So

X = 990

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

By the Central Limit Theorem

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

$Z = \frac{990 - 975}{9.34}$

$Z = 1.61$

$Z = 1.61$ has a pvalue of 0.9463

X = 960

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

$Z = \frac{960 - 975}{9.34}$

$Z = -1.61$

$Z = -1.61$ has a pvalue of 0.0537

0.9463 - 0.0537 = 0.8926

0.8926 = 89.26% probability that the mean weight of the sample of horses would differ from the population mean by less than 15lbs if 31 horses are sampled at random from the stable

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3 years ago