Answer:
the null and alternative hypothesis are
: p=0.24
: p≠0.24
Test statistic is 1.416.
P-Value is 0.1568
We fail to reject the null hypothesis under 0.05 significance level
<em>At 0.05 significance</em> we cannot say that the proportion of offspring peas will be yellow is not 0.24 (24%)
Step-by-step explanation:
Let p be the proportion of offspring peas will be yellow. Then,
: p=0.24
: p≠0.24
test statistic can be calculated as
z=
where
- p(s) is the sample proportion of yellow offspring peas (

- p is the proportion of yellow offspring peas assumed under null hypothesis. (0.24)
- N is the sample size (585)
Test statistic is z=
≈ 1.416.
Using z-table, we can find that P-Value is 0.1568
We fail to reject the null hypothesis under 0.05 significance level since 0.157>0.05
We can conclude that <em>at 0.05 significance</em> we cannot say that the proportion of offspring peas will be yellow is not 0.24 (24%)
620 --> aumentando 35%
35% de 620 = 35*620/100 = 35*62/10 = 217 (aumento)
Montante = 620 + 217 = 837
If there are 48 tiles in each row then minus 48 - 6 white tiles = 42 purple tiles. And if there are 8 rows then 8 x 42 = 336. So Abigail needs 336 purple tiles.
hoped I helped
Answer:
1. 60.50(0.20)
2. 34(0.15) <How much the coupon takes off
3. 157.30(0.10) <How much the coupon takes off
4. 10.49/58.25 <The percentage of the tip
5. 94.20(0.20) <The percentage of the tip
94.20(0.15) <How much the coupon takes off of the price of the meal.
<h3>
Hope I helped!</h3>