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julsineya [31]
3 years ago
6

Mike traveled from the school to the mall and then from the mall to the library. Henry traveled from the school to the library.

How many more miles did Mike travel than Henry?
Mathematics
1 answer:
Natalija [7]3 years ago
4 0

Answer:

12 miles

Step-by-step explanation:

From the graph given :

Mike's travel :

School to Mall= 16 miles

Mall to Library = 30 miles

Total travel = 16 + 30 = 46 miles

Henry's travel :

From school to Library = Hypotenus :

Distance can be obtained using Pythagoras :

Hypotenus = √(opposite² + adjacent²)

Hypotenus = √(30² + 16²)

Hypotenus = √1156

Hypotenus = 34 miles

Difference in Distance covered :

(46 miles - 34 miles) = 12 miles

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olya-2409 [2.1K]

Answer:

<em>DF = 10 units</em>

m\angle DFG = 28^\circ

<em>EG = 5.04</em>

Step-by-step explanation:

<u>Properties of Rhombus es</u>

  1. All Sides Of The Rhombus Are Equal.
  2. The Opposite Sides Of A Rhombus Are Parallel.
  3. Opposite Angles Of A Rhombus Are Equal.
  4. In A Rhombus, Diagonals Bisect Each Other At Right Angles.
  5. Diagonals Bisect The Angles Of A Rhombus.

The image contains a rhombus with the following data (assume the center as point O):

DO = 5 units

GF = 5.6 units

m\angle FEO = 62^\circ

4. Calculate DF

Applying property 4, diagonals bisect each other, thus the length of DF is double the length of DO, i.e. DF=2*5 = 10:

DF = 10 units

5. Calculate m\angle DFG

Applying property 4 in triangle EFO, the center angle is 90°, thus angle EFO has a measure of 90°-62°=28°.

Applying property 5, this angle is half of the measure of angle EFG and angle DFG has the same measure of 28°.

m\angle DFG = 28^\circ

6. FG is the hypotenuse of triangle OFG, thus:

OG^2=FG^2-OF^2

OG^2=5.6^2-5^2

OG^2=6.36

OG=\sqrt{6.36}=2.52

EG is double OG: OG=2*2.52=5.04

EG = 5.04

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2 years ago
If in a sample of 355 adult males, we have a mean total cholesterol level of 185 mg, with s = 16. What is the 95% confidence int
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Answer:

Step-by-step explanation:

We want to determine a 95% confidence interval for the mean total cholesterol level of all males.

Number of sample, n = 355

Mean, u = 185 mg

Standard deviation, s = 16

For a confidence level of 95%, the corresponding z value is 1.96. This is determined from the normal distribution table.

We will apply the formula

Confidence interval

= mean +/- z ×standard deviation/√n

It becomes

185 +/- 1.96 × 16/√355

= 185 +/- 1.96 × 0.849

= 185 +/- 1.66404

The lower end of the confidence interval is 185 - 1.66404 =183.336

The upper end of the confidence interval is 185 + 1.66404 = 186.66

Therefore, with 95% confidence interval, the mean total cholesterol level of all males is between 183.336 mg and 186.66 mg

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Answer:

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Step-by-step explanation:


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