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Zinaida [17]
3 years ago
13

Which decimal is equivalent to 180⁄333? A. 1.24 B. 1.240 C. 1.24024 D. 1.240

Mathematics
2 answers:
Daniel [21]3 years ago
7 0
The correct answer is 0.541 but this wasn't a choice. Hope this helped you. Have a nice day :)
Elis [28]3 years ago
6 0

Answer:

The correct answer is 0.541, however it isn't a choice you provided, hope that is still able to help you in some way!

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Can someone help me with this question please
Elodia [21]
It's the second one -7x^2 because a quadratic equation needs to be squared and that's the only one that provides that, you can also find the differences of each equation to see if it's quadratic
7 0
3 years ago
Read 2 more answers
Help i'll mark brainliest
tensa zangetsu [6.8K]
The correct answer is D. 40
4 0
2 years ago
Read 2 more answers
-1/2+c=31/4<br><br> A. C=33/4<br> B. C=8<br> C. 29/4<br> D. 7
kkurt [141]

Answer:

A

Step-by-step explanation:

-1/2+c=31/4

-1/2+c+1/2=31/4+1/2

c=33/4

4 0
3 years ago
The mean life of a television set is 119119 months with a standard deviation of 1414 months. If a sample of 7474 televisions is
Pepsi [2]

Answer:

0.5034 = 50.34% probability that the sample mean would differ from the true mean by less than 1.1 months

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 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.

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

In this problem, we have that:

\mu = 119, \sigma = 14, n = 74, s = \frac{14}{\sqrt{74}} = 1.6275

What is the probability that the sample mean would differ from the true mean by less than 1.11 months?

This is the pvalue of Z when X = 119 + 1.1 = 120.1 subtracted by the pvalue of Z when X = 119 - 1.1 = 117.9. So

X = 120.1

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

By the Central Limit Theorem

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

Z = \frac{120.1 - 119}{1.6275}

Z = 0.68

Z = 0.68 has a pvalue of 0.7517

X = 117.9

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

Z = \frac{117.9 - 119}{1.6275}

Z = -0.68

Z = -0.68 has a pvalue of 0.2483

0.7517 - 0.2483 = 0.5034

0.5034 = 50.34% probability that the sample mean would differ from the true mean by less than 1.1 months

3 0
3 years ago
Kinda confused on what the sum would be?!? HELP NOWW PLEASE
valentinak56 [21]
That is b. aka 2
thanks
4 0
3 years ago
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