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3 years ago

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What percent of 37 is 111?

1 answer:

exis [7]3 years ago

4 0

111 is 300% of 37. This is how I do it. 37x=111 x=111/37 x=3 move the decimal point right 2 places to get the percentage 300% ANSWER 300%

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Difference in the following polynomials (6x^3 - 2x^2 + 4)<br> -(2x^3 + 4x^2 - 5)

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$\bold{[ \ Answer \ ]}$

$\boxed{\bold{4x^3-6x^2+9}}$

$\bold{[ \ Explanation \ ]}$

$\bold{Find \ Difference: \ \left(6x^3\:-\:2x^2\:+\:4\right)-\left(2x^3\:+\:4x^2\:-\:5\right)}$

$\bold{-----------------------}$

$\bold{Remove \ Parenthesis}$

$\bold{6x^3-2x^2+4-\left(2x^3+4x^2-5\right)}$

$\bold{Remove \ Parenthesis \ / \ Simplify}$

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$\bold{Rewrite}$

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3 years ago

An Article a Florida newspaper reported on the topics that teenagers most want to discuss with their parents. The findings, the

Sergio039 [100]

Answer:

The estimate is $P__{hat}} \pm E = 0.37 \pm 0.0348$

Step-by-step explanation:

From the question we are told that

The sample size is n = 522

The sample proportion of students would like to talk about school is $\r p__{hat}} = 0.37$

Given that the confidence level is 90 % then the level of significance can be mathematically evaluated as

$\alpha = 100 - 90$

$\alpha = 10\%$

$\alpha = 0.10$

Next we obtain the critical value of $\frac{\alpha }{2}$ from the normal distribution table, the value is

$Z_{\frac{\alpha }{2} } =Z_{\frac{0.10}{2} } = 1.645$

Generally the margin of error can be mathematically represented as

$E = Z_{\frac{\alpha }{2} } * \sqrt{\frac{\r P_{hat}(1- \r P_{hat} )}{n } }$

=> $E = 1.645 * \sqrt{\frac{0.37 (1- 0.37 )}{522 } }$

=> $E = 0.0348$

Generally the estimate the proportion of all teenagers who want more family discussions about school at 90% confidence level is

$P__{hat}} \pm E$

substituting values

$0.37 \pm 0.0348$

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3 years ago