Answer:
10 tiles
2 vowel (E, I)
Then P(pick vowel) = 2/10 =1/5 =20%
is simply the difference of both amounts, but firstly let's convert the mixed fractions to improper, and subtract.
![\bf \stackrel{mixed}{4\frac{1}{2}}\implies \cfrac{4\cdot 2+1}{2}\implies \stackrel{improper}{\cfrac{9}{2}} \\\\\\ \stackrel{mixed}{6\frac{7}{16}}\implies \cfrac{6\cdot 16+7}{16}\implies \stackrel{improper}{\cfrac{103}{16}} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ \stackrel{Jessie}{\cfrac{103}{16}}-\stackrel{Bryce}{\cfrac{9}{2}}\implies \stackrel{\textit{our LCD is 16}}{\cfrac{(1)103-(8)9}{16}}\implies \cfrac{103-72}{16}\implies \cfrac{31}{16}\implies 1\frac{15}{16}](https://tex.z-dn.net/?f=%5Cbf%20%5Cstackrel%7Bmixed%7D%7B4%5Cfrac%7B1%7D%7B2%7D%7D%5Cimplies%20%5Ccfrac%7B4%5Ccdot%202%2B1%7D%7B2%7D%5Cimplies%20%5Cstackrel%7Bimproper%7D%7B%5Ccfrac%7B9%7D%7B2%7D%7D%0A%5C%5C%5C%5C%5C%5C%0A%5Cstackrel%7Bmixed%7D%7B6%5Cfrac%7B7%7D%7B16%7D%7D%5Cimplies%20%5Ccfrac%7B6%5Ccdot%2016%2B7%7D%7B16%7D%5Cimplies%20%5Cstackrel%7Bimproper%7D%7B%5Ccfrac%7B103%7D%7B16%7D%7D%0A%5C%5C%5C%5C%5B-0.35em%5D%0A%5Crule%7B34em%7D%7B0.25pt%7D%5C%5C%5C%5C%0A%5Cstackrel%7BJessie%7D%7B%5Ccfrac%7B103%7D%7B16%7D%7D-%5Cstackrel%7BBryce%7D%7B%5Ccfrac%7B9%7D%7B2%7D%7D%5Cimplies%20%5Cstackrel%7B%5Ctextit%7Bour%20LCD%20is%2016%7D%7D%7B%5Ccfrac%7B%281%29103-%288%299%7D%7B16%7D%7D%5Cimplies%20%5Ccfrac%7B103-72%7D%7B16%7D%5Cimplies%20%5Ccfrac%7B31%7D%7B16%7D%5Cimplies%201%5Cfrac%7B15%7D%7B16%7D)
Answer:
As a decimal its 6.34, If you round up then its 6.35
As a fraction its 400/63
As a mixed number its 6 22/63
Answer:
The P-value is 0.0166.
Step-by-step explanation:
<u>The complete question is:</u> In a one-tail hypothesis test where you reject H0 only in the lower tail, what is the p-value if ZSTAT = -2.13.
We are given that the z-statistics value is -2.13 and we have to find the p-value.
Now, the p-value of the test statistics is given by the following condition;
P-value = P(Z < -2.13) = 1 - P(Z
2.13)
= 1 - 0.9834 = <u>0.0166</u>
Assuming that the level of significance is 0.10 or 10%.
The decision rule for rejecting the null hypothesis based on p-value is given by;
- If the P-value of the test statistics is less than the level of significance, then we have sufficient evidence to reject the null hypothesis.
- If the P-value of the test statistics is more than the level of significance, then we have insufficient evidence to reject the null hypothesis.
Here, the P-value is more than the level of significance as 0.0166 > 0.10, so we have insufficient evidence to reject the null hypothesis, so we fail to reject the null hypothesis.