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adell [148]
3 years ago
14

Between 6 p.m. and 10 a.m. the temperature Rose by 5 degrees Fahrenheit between 10 a.m. and 2 p.m. the temperature Rose by 6 deg

rees Fahrenheit between 2 p.m. and 6 p.m. the temperature Rose by 0 Celsius. Would you be able to tell how much the temperature rose since 6 a.m? Why or why not?​
Mathematics
1 answer:
Art [367]3 years ago
3 0

Answer:

you got this your smart

Step-by-step explanation:

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Naresh walked 2 km 35 m in the morning and 1 km 7 m in the evening. How much distance did he walk in all?
lara [203]
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4 0
3 years ago
A trampoline has an area of 49pi square feet what's the diameter of the trampoline
Anna007 [38]
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7 0
3 years ago
A data set lists weights​ (lb) of plastic discarded by households. The highest weight is 5.31 ​lb, the mean of all of the weight
a_sh-v [17]

Answer:

a) The difference is of 3.222 lbs.

b) 1.64 standard deviations.

c) Z = 1.64

d) Not significant, as the z-score of 1.64 is between -2 and 2.

Step-by-step explanation:

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the z-score 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 p-value, we get the probability that the value of the measure is greater than X.

The mean of all of the weights is x=2.088 ​lb, and the standard deviation of the weights is s=1.968 lb.

This means that \mu = 2.088, \sigma = 1.968

a. What is the difference between the weight of 5.31 lb and the mean of the​ weights?

This is X - \mu = 5.31 - 2.088 = 3.222

The difference is of 3.222 lbs.

b. How many standard deviations is that​ [the difference found in part​ (a)]?

This is the z-score. So

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

Z = \frac{5.31 - 2.088}{1.968}

Z = 1.64

1.64 standard deviations.

c. Convert the weight of 5.31 lb to a z score.

Z = 1.64, as found above.

d. If we consider weights that convert to z scores between −2 and 2 to be neither significantly low nor significantly​ high, is the weight of 5.31 lb​ significant?

Not significant, as the z-score of 1.64 is between -2 and 2.

5 0
3 years ago
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