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aleksandrvk [35]
3 years ago
12

David and Mark are marking exam papers. Each set takes David 24 minutes and Mark 1 hour. Express the times David and Mark take a

s a ratio.
Give your answer in its simplest form
Mathematics
1 answer:
attashe74 [19]3 years ago
8 0

Answer:

2:5

Step-by-step explanation:

david=24mins

mark = 1hour

change 1 hour to mins

DAVID=24

MARK=60

24:60

THEN SIMPLIFY

24/60

<em>=</em><em>2</em><em>/</em><em>5</em>

<em><u>2</u></em><em><u>:</u></em><em><u>5</u></em>

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Can someone walk me through how to do this problem please?​
Sholpan [36]

Hey!

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We Know:

m∠AED = 34°

m∠EAD = 112°

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Solution:

You notice 4 small triangles in both triangles. That shows that both triangles are the same.

The angles are the same for m∠BDC and m∠AED.

The angles are the same for m∠ADB and m∠EAD

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Angles:

m∠BDC = 34°

m∠ADB = 112°

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Congruent Angles:

m∠AED ≡ m∠BDC

m∠EAD ≡ m∠ADB

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Hope This Helped! Good Luck!

5 0
3 years ago
What is the integral of a square root? Please explain.
IrinaK [193]
If y = xⁿ

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y = √x

y = x^(0.5)

∫y dx  =  x^(0.5+1) / (0.5 + 1) =   x^(1.5)  / 1.5 = x^(1.5)  / (3/2)

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∫y dx = (2/3) x^(3/2)  + C.

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5 0
3 years ago
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You arrive at an intersection with traffic lights that are not working because of a power outage
yulyashka [42]

Traffic rules are set by each individual state.

In most states, an intersection where traffic lights are out of order
is treated as if it were a 4-way STOP.

But you can't depend on this answer.  You MUST determine what
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5 0
3 years ago
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Golf-course designers have become concerned that old courses are becoming obsolete since new technology has given golfers the ab
zhenek [66]

Answer:

test statistic is ≈ -0.36

p-value is  ≈ 0.64

There is no significant evidence that the average golfer can hit the ball more than 235 yards on average.

Step-by-step explanation:

a hypothesis test where H_0: mu = 235 and H_1:mu > 235

test statistic can be calculated as follows:

z=\frac{X-M}{\frac{s}{\sqrt{N} } } where

  • sample mean driving distance (233.8 yards)
  • M is the average expected distance that the average golfer can hit the ball under null hypothesis. (235 yards)
  • s is the standard deviation (46.6 yards)
  • N is the sample size (192)

Then test statistic is z=\frac{233.8-235}{\frac{46.6}{\sqrt{192} } } =-0.3568

p-value is  0.64 >0.05

There is no significant evidence that the average golfer can hit the ball more than 235 yards on average.

6 0
4 years ago
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Answer:

376.8 ft²

Step-by-step explanation:

that is the correct answer for you friend

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3 years ago
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