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2 years ago

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18/5 divided by 3/25

id="TexFormula1" title=" \frac{18}{5} \div \frac{3}{25} " alt=" \frac{18}{5} \div \frac{3}{25} " align="absmiddle" class="latex-formula">

2 answers:

Rudik [331]2 years ago

8 0

Answer:

30

Step-by-step explanation:

There are a few different ways we can approach this problem.

The easiest way is to flip the second fraction and multiply:

$\dfrac{18}{5} \div \dfrac{3}{25}=\dfrac{18}{5} \times \dfrac{25}{3}=\dfrac{18 \times 25}{5 \times 3}$

To do this without a calculator, rewrite 18 as 6 x 3 and 25 as 5 x 5:

$\dfrac{18 \times 25}{5 \times 3}=\dfrac{6 \times 3 \times 5 \times 5}{5 \times 3}$

Now we can cancel out the common factors of 5 and 3 from the numerator and denominator, and are left with:

$\implies 6 \times 5 =30$

brilliants [131]2 years ago

4 0

Answer:

30 or 30/1

Step-by-step explanation:

keep flip change

keep the first fraction the same, the second fraction Flipped and change the sign

18×25

5×3

=450/15

=30

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3 years ago

Abnormalities In the 1980s, it was generally believed that congenital abnormalities affected about 5% of the nation’s chil-dren.

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Answer:

Null hypothesis: $p\leq 0.05$

Alternative hypothesis: $p>0.05$

The conditions and requirements are explained on detail below.

Step-by-step explanation:

1) Data given and notation n

n=384 represent the random sample taken

X=46 represent the children with abnormalities in the sample

$\hat p=\frac{46}{384}=0.120$ estimated proportion of children with abnormalities in the sample

$p_o=0.05$ is the value that we want to test

$\alpha$ represent the significance level (no given)

z would represent the statistic (variable of interest)

$p_v$ represent the p value (variable of interest)

p= proportion of children with congenital abnormalities

2) Concepts and formulas to use

We need to conduct a hypothesis in order to test the claim that the proportion of children with congenital abnormalities exceeds 5%. :

Null hypothesis: $p\leq 0.05$

Alternative hypothesis: $p>0.05$

We assume that the proportion follows a normal distribution.

When we conduct a proportion test we need to use the z statistic, and the is given by:

$z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}$ (1)

The One-Sample Proportion Test is used to assess whether a population proportion $\hat p$ is significantly (different,higher or less) from a hypothesized value $p_o$ .

<em>Check for the assumptions that he sample must satisfy in order to apply the test</em>

a)The random sample needs to be representative: On this case the problem no mention about it but we can assume it.

b) The sample needs to be large enough

$np_o =384*0.05=19.2>10$

$n(1-p_o)=384*(1-0.05)=364.8>10$

3) Calculate the statistic

Since we have all the info requires we can replace in formula (1) like this:

$z=\frac{0.120 -0.05}{\sqrt{\frac{0.05(1-0.05)}{384}}}=6.29$

4) Statistical decision

It's important to refresh the p value method or p value approach . "This method is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.

The significance level is not provided usually is $\alpha=0.05$ . The next step would be calculate the p value for this test.

Since is a one side test the p value would be:

$p_v =P(z>6.29)=1.59x10^{-10}$

Based on the p value obtained and using the significance level given $\alpha=0.05$ we have $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the proportion of children with abnormalities exceeds 0.05 or 5% .

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4 years ago