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2 years ago

Show 46/200 on the hundredths grid. then write the fractions as a demical

1 answer:

3 0

If we want to has 100 in the denominator we have to divide numerator and denominator by 2. Then total value of fraction doesnt change.
$\frac{46}{200}= \frac{46:2}{200:2}= \frac{23}{100}$
Now read the fraction, its 23 hundreds, now just write as decimal, so, 0.23

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2 years ago

According to the US Bureau of labor statistics, 7% of US female workers between 16 and 24 years old are paid at the minimum wage

vladimir1956 [14]

Answer:

The Null Hypothesis is $H_o:k_o = 0.07$

The alternative hypothesis is $H_a :k_o \ne 0.07$

Decision rule

If the test staistics is greater than the critical value of significance level then $H_o$ is accepted else $H_o$ is rejected

With the above in mind

The critical value of the significance level which is obtained from the table is

$t_{0.05} = 1.645$

Now since the critical value of significance level is greater than the test staistics then the null hypothesis will be rejected

Conclusion

The information is not enough to back the claim that state differs from the nation

Step-by-step explanation:

From the question we are told that

The percentage of US female workers paid at the minimum wage or less is $k_o =$ 7% = 0.07

The sample size is $n = 500$

The number paid minimum wage or less is x = 42

The significance level is $\alpha =$ 5% = 0.05

Now the probability of getting a US female workers paid at the minimum wage or less is mathematically represented as

$\= k = \frac{x}{n}$

substituting value

$\= k = \frac{42}{500}$

$\= k = 0.084$

The Null Hypothesis is $H_o:k_o = 0.07$

The alternative hypothesis is $H_a :k_o \ne 0.07$

Generally the test statistics is mathematically evaluated as

$z = \frac{\= k - k_o}{\sqrt{\frac{k_o(1-k_o)}{n} } }$

substituting value

$z = \frac{0.084 - 0.07}{\sqrt{\frac{0.07 (1-0.07)}{500} } }$

$z = 1.23$

Now the Decision rule is stated as

If the test staistics is greater than the critical value of significance level then $H_o$ is accepted else $H_o$ is rejected

With the above in mind

The critical value of the significance level which is obtained from the table is

$t_{0.05} = 1.645$

Now since the critical value of significance level is greater than the test staistics then the null hypothesis will be rejected

So the conclusion will be

The information is not enough to back the claim that state differs from the nation

4 0

2 years ago