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LiRa [457]
3 years ago
12

Suppose point z is halfway between point w and x a standard numberline, and point x is halfway between points z and y. Where is

w if z is located at 1/3and y is located at 11/3

Mathematics
1 answer:
Delicious77 [7]3 years ago
6 0

Answer:

The point w is located at -4/3

Step-by-step explanation:

Point z is halfway between point w and x

Point x is halfway between point z and y

Point z is located at 1/3 and point y is located at 11/3

Please refer to the number line attached

Then the x is located at the mid-point of z and y

x = \frac{z+y}{2} \\\\x = \frac{\frac{1}{3}+ \frac{11}{3} }{2}

x = 2

Since z is the mid-point of w and x

z = \frac{w+ x }{2}\\\\\frac{1}{3}  = \frac{w+ 2 }{2}\\\\\frac{2}{3}  = w+ 2 \\\\w = \frac{2}{3} - 2\\\\w = -\frac{4}{3}

Therefore, the point w is located at -4/3

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So we need to convert 0.75 feet to inches.

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6 0
3 years ago
Read 2 more answers
A recent survey of 128 high school students indicated the following information about release times from school.
Delicious77 [7]

Answer:

A) Interval estimate for those that want to get out earlier = (35%) ± (4%) = (31%, 39%)

Interval estimate for those that want to get out later = (39%) ± (4%) = (35%, 43%)

B) The group that wants to get out of school earlier can win after all the votes are counted if their true population proportion takes on a value that is higher than the closest true population proportion (for the group that wants to get out of school later)

That is, in the (31%, 39%) and (35%, 43%) obtained in (a), a range of (35.1%, 39%) and (35%, 38.9%) show how possible that the group that wants to get out of school earlier can win after all the votes are counted.

Step-by-step explanation:

The Interval estimate for the proportion is basically an interval of range of values where the true population proportion can be found with a certain level of confidence.

Mathematically,

Interval estimate = (Sample proportion) ± (Margin of error)

Margin of Error is the width of the confidence interval about the proportion.

a) Find the interval estimates

i) for those that want to get out earlier and

ii) those that want to get out later.

i) Sample proportion of those that want to get out earlier = 35%

Margin of Error = 4%

Interval estimate for those that want to get out earlier = (35%) ± (4%) = (31%, 39%)

ii) Sample proportion of those that want to get out later = 39%

Margin of Error = 4%

Interval estimate for those that want to get out later = (39%) ± (4%) = (35%, 43%)

b) Explain how it would be possible for the group that wants to get out of school earlier to win after all the votes are counted.

Since the interval estimates represent the range of values that the true population proportion can take on for each group that prefer a particular option, the group that wants to get out of school earlier van have their proportion take on values between 31% and 39%. If their true population takes on a value that is highest (which is very possible from the interval estimate), and the group with the highest proportion in the sample, (the group that wants to get out of school later, whose true population proportion can take between 35% and 43%) has a true population proportion that is less than that of the group that wants to get out of school earlier, then, the group that wants to get out of school earlier can win after all the votes are counted.

Hope this Helps!!!

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3 years ago
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Step-by-step explanation:

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3 0
2 years ago
Solve for h<br> -8 &lt; -4h + 20
Vladimir [108]

Answer:

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Step-by-step explanation:

-8 < -4h + 20

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