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3 years ago

Which number is between 1/6 and 0.25?

Mathematics

1 answer:

Airida [17]3 years ago

5 0

$\displaystyle \frac{1}{6} = 0.16 \\ \\ 0.16 , \boxed{0.17 ,0.18,0.19,0.20,0.21,0.22,0.23,0.24},0.25$

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Serhud [2]

The answer is 3 becasue

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3 years ago

Read 2 more answers

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andrew11 [14]

(b) is the answer.

Step-by-step explanation:

By the Pythagorean Theorem,

A² + B² = C²

Where:

A = Length of side 1

B = Length of side 2

C = Hypotenuse

This rule applies to all right-angled triangles.

The length of the hypotenuse of a right-angled triangle is always the largest value.

Therefore, we can test the answers with the equation above.

(a)

8² + 18² = 20²

64 + 324 = 400

388 ≠ 400

The rule of Pythagorean theorem doesn't work on a, so (a) is not a right-angled triangle.

(b)

12² + 35² = 37²

144 + 1225 = 1369

1369 = 1369

The rule of Pythagorean theorem works here, so (b) is a right-angled triangle.

8 0

3 years ago

4, 8, 12, 16, 20 24, 36, 40 44 48 52 56 60 64 68 72 76 80 84 88 92 96 100... Those are the multiples of 4. Now find out which ar

Lilit [14]

Answer:

Ok the answer of 6 are 12, 18, 24, 30, 36, 42, and 48. The multiples of 9 are 18,27,36,45,54,63,72,81,90,99,108,117,126,135,144,153,162,171,180,189,198,. Hope it helps!

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3 years ago

A 500 g model train car traveling at 0.8 m/s collides with a 300 g stationary car. The cars hook up and move off down the track

sdas [7]

Answer:

$0.5ms^{-1}$

Step-by-step explanation:

Let v be the speed after the collision of both the cars.

Now, momentum is given by: mass × velocity, then by equating the total momentum before and after the collision of the two cars, we get

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⇒ $400=800v$

⇒ $v=0.5ms^{-1}$

Thus, they are moving at the velocity of $0.5ms^{-1}$

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3 years ago

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on a large college campus, 84% of the students report drinking alcohol within the past month, 33% report using some type of toba

Natasha_Volkova [10]

Answer: 0.31

Step-by-step explanation:

Let A denotes the event that the students report drinking alcohol and B denotes the students report using some type of tobacco product .

Given : P(A) =0.84 ; P(B)=0.33 and P(A∪B)=0.86

We know that $P(A\cap B)=P(A)+P(B)-P(A\cup B)$

Then, the probability that the student both drunk alcohol and used tobacco in the past month is given by :-

$P(A\cap B)=0.84+0.33-0.86=0.31$

Hence, the probability that the student both drunk alcohol and used tobacco in the past month = 0.31

5 0

3 years ago