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WARRIOR [948]
3 years ago
14

Which amount is greater, 20 mm or 1 cm?

Mathematics
1 answer:
Semmy [17]3 years ago
3 0

Answer:

20mm

Step-by-step explanation: 1cm is equal to 10mm

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Find the values of x for which the denominator is equal to zero for y=3x+5/x-6.
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\huge\color{black}\underline\mathbb{\:ANSWER}

\green{••••••••••••••••••••••••••••••••••••••••••••••••••}

Find the values of x for which the denominator is equal to zero for y=3x+5/x-6.

x = −5

x = 6

<h2>x = −6</h2>

therefore the answer is

<h2>x = −6</h2>

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2 years ago
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Best way to tell someone no without being mean
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Try to search up euphemisms to help you in an argument where your forced to say no

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I can't do that or no

6 0
3 years ago
9:45 on Tuesday Questions
nirvana33 [79]

Answer:

Alternate Exterior Angles

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The alternate exterior angle theorem is when two different lines are corssed by a transversal. Essentially alternate exterior angles are when two different angles are on opposite sides of the transversal. In the given photo, 6 and 12 are on opposite sides of one transversal which makes them alternate exterior angles.

Best of Luck!

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A meteorologist who sampled 4 thunderstorms found that the average speed at which they traveled across a certain state was 16 mi
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Answer:

The  90 % confidence  interval  for the mean population is (11.176  ; 20.824 )

Rounding to at least two decimal places would give 11.18 , 20.83

Step-by-step explanation:

Mean = x`= 16 miles per hour

standard deviation =s= 4.1 miles per hour

n= 4

\frac{s}{\sqrt n}  =  4.1/√4= 4.1/2= 2.05

1-α= 0.9

degrees of freedom =n-1=  df= 3

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Using Students' t - test

x`±∈ * \frac{s}{\sqrt n}

Putting values

16 ± 2.353 * 2.05

= 16 + 4.82365

20.824  ;        11.176

The  90 % confidence  interval  for the mean population is (11.176  ; 20.824 )

Rounding to at least two decimal places would give 11.18 , 20.83

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3 years ago
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A 8 ft tall cardboard box standing next to a man casts a 20 ft shadow. If the man casts a shadow that is 15 ft long, then how ta
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The answer is that he would be 27 ft tall

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