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tia_tia [17]
2 years ago
12

Reggie has $350 he received for his birthday. Every month, $25 is removed from his account to go to his college savings (no othe

r transactions are made). Create and solve an equation that can be used to find the number of months, m, it takes for the account balance to reach $225.
I need an equation ASAP!
Mathematics
2 answers:
BartSMP [9]2 years ago
4 0

Answer:

reggie is stuopid he should spend it on bubgle gum.

Step-by-step explanation:

professor190 [17]2 years ago
4 0

Answer:

225=-25x+350 (x=5 months)

Step-by-step explanation:

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Researchers measured the data speeds for a particular smartphone carrier at 50 airports. The highest speed measured was 75.6 Mbp
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a) 59.98

b) 2.99

c) 2.99

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Part a)

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Average/Mean speed = \overline{x} = 15.62 Mbps

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x - \overline{x} = 75.6 - 15.62 = 59.98 Mbps

Thus, the difference between​ carrier's highest data speed and the mean of all 50 data​ speeds is 59.98 Mbps

Part b)

In order to find how many standard deviations away is the difference found in previous part, we divide the difference by the value of standard deviation i.e.

\frac{59.98}{20.03}=2.99

This means, the difference is 2.99 standard deviations or in other words we can say, the Carrier's highest data speed is 2.99 standard deviations above the mean data speed.

Part c)

A z score tells us that how many standard deviations away is a value from the mean. We calculated the same in the previous part. Performing the same calculation in one step:

The formula for the z score is:

z=\frac{x-\overline{x}}{s}

Using the given values, we get:

z=\frac{75.6-15.62}{20.03}=2.99

Thus, the Carriers highest data is equivalent to a z score of 2.99

Part d)

The range of z scores which are neither significantly low nor significantly​ high is -2 to + 2. The z scores outside this range will be significant.

Since, the z score for carrier's highest data speed is 2.99 which is well outside the given range, i.e. greater than 2, we can conclude that the  carrier's highest data speed​ is significantly higher.

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