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12345 [234]
3 years ago
11

Divide. Write in simplest form. 10 divided by 1 2/3 ?

Mathematics
1 answer:
adell [148]3 years ago
8 0

Answer: 6

Step-by-step explanation: First rewrite 10 as 10/1 and 1 and 2/3 as 5/3.

Mixed numbers can be changed to improper fractions by multiplying the denominator by the whole number and then adding the numerator. We then put out numerator over our old denominator.

So we have 10/1 ÷ 5/3 or 10/1 × 3/5.

It's important to understand that dividing by a fraction is the same as multiplying by its reciprocal. In other words, we can change the division to multiplication and flip the second fraction.

Now multiplying across the numerators and across the denominators, we have 30/5. Notice however that 30/5 is not in lowest terms so we divide the numerator and the denominator by the greatest common factor of 30 and 5 which is 6 and we end up with 6.

Therefore, 10 ÷ 1 and 2/3 = 6.

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The weights of 67 randomly selected axles were found to have a variance of 3.85. Construct the 80% confidence interval for the p
VLD [36.1K]

Answer:

3.13

Step-by-step explanation:

We are given the following in the question:

Sample size, n = 67

Variance = 3.85

We have to find 80% confidence interval for the population variance of the weights.

Degree of freedom = 67 - 1 = 66

Level of significance = 0.2

Chi square critical value for lower tail =

\chi^2_{1-\frac{\alpha}{2}}= 51.770

Chi square critical value for upper tail =

\chi^2_{\frac{\alpha}{2}}= 81.085

80% confidence interval:

\dfrac{(n-1)S^2}{\chi^2_{\frac{\alpha}{2}}} < \sigma^2 < \dfrac{(n-1)S^2}{\chi^2_{1-\frac{\alpha}{2}}}

Putting values, we get,

=\dfrac{(67-1)3.85}{81.085} < \sigma^2 < \dfrac{(67-1)3.85}{51.770}\\\\=3.13

Thus, (3.13,4.91) is the required 80% confidence interval for the population variance of the weights.

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3 years ago
Please Help ASAP
Tju [1.3M]

9514 1404 393

Answer:

  no solution

Step-by-step explanation:

The discriminant is ...

  b^2 -4ac = (-4)^2 -4(2)(16) = 16 -128 = -112

A negative discriminant means there are no real solutions to the equation.

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The complex solutions are 1±i√7.

If we assume the equation has a typographical error and is intended to be ...

  2x^2 -4x -16 = 0

The solutions are x = -2 and x = 4.

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3 years ago
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