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marta [7]
2 years ago
11

The mean high temperature on a particular day in January is 31 degrees F and the standard deviation is 8.7 degrees. One year, th

e temperature was 52 degrees F on that day. What is the Z-score for that day's temperature? Round to the appropriate number of decimal places for Z-scores. Is this temperature significantly high? (greater than 2)​
Mathematics
1 answer:
Basile [38]2 years ago
8 0

Answer:

The Z-score for that day's temperature is 2.41, and since Z > 2, this temperature is significantly high.

Step-by-step explanation:

Z-score:

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.

If |Z| > 2, the measure X is significantly high(Z > 2) or significantly low(Z < -2).

The mean high temperature on a particular day in January is 31 degrees F and the standard deviation is 8.7 degrees.

This means that \mu = 31, \sigma = 8.7

One year, the temperature was 52 degrees F on that day.

This means that X = 52.

What is the Z-score for that day's temperature?

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

Z = \frac{52 - 31}{8.7}

Z = 2.41

The Z-score for that day's temperature is 2.41, and since Z > 2, this temperature is significantly high.

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Answer:

Volume of gumball without including its hollow core is 48347.6 cubic mm.

Step-by-step explanation:

Given:

Diameter of Gumball = 66 mm

Since radius is half of diameter.

Radius of gumball = \frac{diameter}{2}=\frac{66}{2} =33 \ mm

Now We will first find the Volume of Gumball.

To find the Volume of Gumball we will use volume of sphere which is given as;

Volume of Sphere = \frac{4}{3}\pi r^3

Now Volume of Gumball = \frac{4}{3}\times3.14 \times (33)^3 = 150456.24 \ mm^3

Also Given

Diameter of gumball's spherical hollow core = 58 mm

Since radius is half of diameter.

Radius of gumball's spherical hollow core = \frac{diameter}{2}=\frac{58}{2} =29 \ mm

Now We will find the Volume of gumball's spherical hollow core.

Volume of Sphere = \frac{4}{3}\pi r^3

So Volume of gumball's spherical hollow core = \frac{4}{3}\times3.14 \times (29)^3 = 102108.61 \ mm^3

Now We need to find volume of the gumball without including its hollow core.

So, To find volume of the gumball without including its hollow core we would Subtract Volume of gumball spherical hollow core from Volume of Gumball.

volume of the gumball without including its hollow core = Volume of Gumball - Volume of gumball's spherical hollow core = 150456.24\ mm^3 - 102108.61\ mm^3 = 48347.63\ mm^3

Rounding to nearest tenth we get;

volume of the gumball without including its hollow core = 48347.6\ mm^3

Hence Volume of gumball without including its hollow core is 48347.6 cubic mm.

8 0
3 years ago
|6x|+3=21 solve for X (absolute value)​
Fudgin [204]

Answer:

x=3\text{ or } x=-3

Step-by-step explanation:

So we have the equation:

|6x|+3=21

First, subtract 3 from both sides:

|6x|=18

Definition of absolute value:

6x=18\text{ or } 6x=-18

Divide both sides by 6 for both equations:

x=3\text{ or } x=-3

And we're done!

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Answer:

How far does Clare travel in one hour?

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How far does Clare travel in three hours?

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