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alexandr402 [8]
3 years ago
11

Estimate using compatible numbers. 53 1/2÷9

Mathematics
2 answers:
Hitman42 [59]3 years ago
6 0

Answer:

Estimate is 6.

Step-by-step explanation:

53 1/2:  Estimate 54.

54 / 9 = 6.

Nitella [24]3 years ago
3 0

Answer:

3

Step-by-step explanation:

53/2 = 26.5 = 27/9 = 3

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A linear equation written in the form y = mx + c is known as,
SCORPION-xisa [38]

Answer:

it is a slope intercept form

7 0
3 years ago
Write .008% as a decimal
lions [1.4K]
0.008%=0.008/100=0.00008
5 0
3 years ago
9. On a game show, there are 16 questions: 8 easy, 5 medium-hard, and 3 hard. If contestants are given questions
just olya [345]

Answer:

\frac{7}{30}\approx0.23

Step-by-step explanation:

Given:

Number of questions (N) = 16

Number of easy questions (E) = 8

Number of medium-hard questions (M) = 5

Number of hard questions (H) = 3

Now, the probability of getting the first question as easy question is given as:

P(E1)=\frac{\textrm{Number of easy questions}}{\textrm{Total questions}}\\\\P(E1)=\frac{E}{N}=\frac{8}{16}=0.5

Now, probability of getting the second question as easy question is given as:

P(E2)=\frac{\textrm{Number of easy questions left}}{\textrm{Total questions left}}\\\\P(E2)=\frac{E-1}{N-1}=\frac{7}{15}

Now, probability that the first two contestants will get easy questions is given by the product of P(E1)\ and\ P(E2). So,

P(2\ easy\ questions)=P(E1)\times P(E2)\\\\P(2\ easy\ questions)=\frac{1}{2}\times \frac{7}{15}\\\\P(2\ easy\ questions)=\frac{7}{30}\approx 0.23

Therefore, the probability that the first two contestants will get easy questions is \frac{7}{30}\approx0.23

7 0
3 years ago
Which expression is a sum of cubes?
erik [133]

we know that

A polynomial in the form a^{3} +b^{3} is called a sum of cubes

so

Let's verify each case to determine the solution

<u>case A)</u> -64x^{6} y^{12} +125x^{16} y^{3}

we know that

-64=-4^{3}

x^{6}= (x^{2})^{3}

y^{12}= (y^{4})^{3}

125=5^{3}

x^{16}=x^{15} *x=x*(x^{5})^{3} -------> is not a perfect cube

y^{3}= (y)^{3}

therefore

the case A) is not a sum of cubes

<u>case B)</u> -32x^{6} y^{12} +125x^{16} y^{3}

we know that

-32=-2^{5} -------> is not a perfect cube

x^{6}= (x^{2})^{3}

y^{12}= (y^{4})^{3}

125=5^{3}

x^{16}=x^{15} *x=x*(x^{5})^{3} -------> is not a perfect cube

y^{3}= (y)^{3}

therefore

the case B) is not a sum of cubes

<u>case C)</u> 32x^{6} y^{12} +125x^{9} y^{3}

we know that

32=2^{5} -------> is not a perfect cube

x^{6}= (x^{2})^{3}

y^{12}= (y^{4})^{3}

125=5^{3}

x^{9}=(x^{3})^{3}

y^{3}= (y)^{3}

therefore

the case C) is not a sum of cubes

<u>case A)</u> 64x^{6} y^{12} +125x^{9} y^{3}

we know that

64=4^{3}

x^{6}= (x^{2})^{3}

y^{12}= (y^{4})^{3}

125=5^{3}

x^{9}=(x^{3})^{3}

y^{3}= (y)^{3}

Substitute

4^{3}(x^{2})^{3}(y^{4})^{3} +5^{3}(x^{3})^{3}(y)^{3}

(4x^{2}y^{4})^{3} +(5x^{3}y)^{3}

therefore

<u>the answer is</u>

64x^{6} y^{12} +125x^{9} y^{3} is a sum of cubes

6 0
3 years ago
Read 2 more answers
Right down all factors of 10
Ugo [173]

Answer:

1, 2, 5, and 10.

Step-by-step explanation:

The factors of 10 are 1, 2, 5, and 10.

2 × 5 = 10

1 × 10 = 10

3 0
3 years ago
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