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iVinArrow [24]
1 year ago
15

The mean amount of money spent on lunch per week for a sample of 100 students is $21. If the margin of error for the population

mean with a 99% confidence interval is 1.50, construct a 99% confidence interval for the mean amount of money spent on lunch per week for all students.
Mathematics
1 answer:
Digiron [165]1 year ago
7 0

From the given mean and margin of error, the 99% confidence interval for the mean amount of money spent on lunch per week for all students is:

[$19.5, $22.5].

<h3>How to calculate a confidence interval given the sample mean and the margin of error?</h3>

The confidence interval is given by the sample mean plus/minus the margin of error, hence:

  • The lower bound is the sample mean subtracted by the margin of error.
  • The upper bound is the sample mean added to the margin of error.

For this problem, we have that:

  • The sample mean is of $21.
  • The margin of error is of $1.50.

Hence the bounds are given as follows:

  • Lower bound: 21 - 1.50 = $19.50.
  • Upper bound: 21 + 1.50 = $22.50.

Hence the interval is [$19.50, $22.50].

More can be learned about confidence intervals at brainly.com/question/25890103

#SPJ1

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Triangle L M Q is cut by perpendicular bisector L N. Angle N L Q is 32 degrees and angle L M N is 58 degrees.
Serga [27]

Answer:

Yes

Step-by-step explanation:

ΔMNL ≅ ΔQNL  by ASA or AAS

by ASA

Proof:

∠ LNM = ∠LNQ    =90

LN = LN   {Common}

∠MLN = ∠QLN     {LN bisects ∠ L}

By AAS

∠Q + ∠QLN + ∠LNQ = 180  {Angle sum property of triangle}

∠Q + 32 + 90 = 180

∠Q  + 122 = 180

∠Q = 180 -122 =

∠Q = 58

∠Q = ∠M

∠MNL =∠QNL = 90

LN = LN {common side}

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