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3 years ago

Pls Someone help me I don't understand math very well can someone help me on how to get the answer for 1/4 x 8 4/9?

Mathematics

2 answers:

pochemuha3 years ago

8 0

Answer:

2 1/9

Step-by-step explanation:

Multiplying fractions: Numerator times numerator over denominator times denominator. IF you have a mixed number, convert it into improper fraction, then do numerator times numerator over denominator times denominator.

1/4 x 76/9 (I got 76/9 because the 8 means 8 9/9s, so (8 x 9)=72 plus 4 because 4/9s=76/9)

1/4 x 76/9= 76/36

=38/18

=19/9

=2 1/9

Eduardwww [97]3 years ago

7 0

Answer:

2 4/36

Step-by-step explanation:

To do this, multiply the whole number by the denominator and add it to the numerator.

8 x 9 = 72

72 + 4 = 76

76/9

Then multiply across (multiply numerators with numerators; denominators with denominators)

76 x 1 = 76

9 x 4 = 36

76/36 = 2 4/36

____________________________

A faster way to do this is to multiply the whole number by the fraction and then the fraction by the fraction.

1/4 x 8/1 = 8/4 = 2

1/4 x 4/9 = 4/36

= 2 4/36

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a) $n=\frac{0.675(1-0.675)}{(\frac{0.02}{1.64})^2}=1475.07$

And rounded up we have that n=1476

b) $n=\frac{0.5(1-0.5)}{(\frac{0.02}{1.64})^2}=1681$

And rounded up we have that n=1681

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The population proportion have the following distribution

$p \sim N(p,\sqrt{\frac{\hat p(1-\hat p)}{n}})$

The margin of error for the proportion interval is given by this formula:

$ME=z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}$ (a)

If solve n from equation (a) we got:

$n=\frac{\hat p (1-\hat p)}{(\frac{ME}{z})^2}$ (b)

Part a

In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 90% of confidence, our significance level would be given by $\alpha=1-0.9=0.1$ and $\alpha/2 =0.05$ . And the critical value would be given by: