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3 years ago

How do i write 12.009 in a word form?

1 answer:

5 0

Your answer will be twelve and nine thousandths because in place value decimals are like tenths, hundredths, thousandths. In place value for regular numbers are ones, tens, hundreds and so on. So, when you have 12.009, you are going to separate the decimals with the numbers. Then, write down the number in words which would be twelve. Now, go to the decimal and write that down, which is 9 thousandths. Finally, combine the both of them to get: twelve and nine thousandths.

Hope this helps :)
and good luck

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a) 0.057

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c) 0.4766

Step-by-step explanation:

To find the p-value if the sample average is 185, we first compute the z-score associated to this value, we use the formula

$z=\frac{\bar x-\mu}{\sigma/\sqrt N}$

where

$\bar x=mean\; of\;the \;sample$

$\mu=mean\; established\; in\; H_0$

$\sigma=standard \; deviation$

N = size of the sample.

So,

$z=\frac{185-175}{20/\sqrt {10}}=1.5811$

$\boxed {z=1.5811}$

As the sample suggests that the real mean could be greater than the established in the null hypothesis, then we are interested in the area under the normal curve to the right of 1.5811 and this would be your p-value.

We compute the area of the normal curve for values to the right of 1.5811 either with a table or with a computer and find that this area is equal to 0.0569 = 0.057 rounded to 3 decimals.

So the p-value is

$\boxed {p=0.057}$

Since the z-score associated to an α value of 0.05 is 1.64 and the z-score of the alternative hypothesis is 1.5811 which is less than 1.64 (z critical), we cannot reject the null, so we are making a Type II error since 175 is not the true mean.

We can compute the probability of such an error following the next steps:

Step 1

Compute $\bar x_{critical}$