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3 years ago

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:

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

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3 years ago

Write .008% as a decimal

lions [1.4K]

0.008%=0.008/100=0.00008

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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

Let's verify each case to determine the solution

case A) $-64x^{6} y^{12} +125x^{16} y^{3}$

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

case B) $-32x^{6} y^{12} +125x^{16} y^{3}$

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

case C) $32x^{6} y^{12} +125x^{9} y^{3}$

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

case A) $64x^{6} y^{12} +125x^{9} y^{3}$

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

the answer is

$64x^{6} y^{12} +125x^{9} y^{3}$ 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