5c + 3 - 2c + 5
5c - 2c + 3 + 5
3c + 8 ~~~
Answer:
4
Step-by-step explanation:
- If you take the lowest half of the data (from the lower extreme to the median) and find the median of these numbers, that is called the lower quartile or quartile 1.
- If you take the upper half of the data (from the upper extreme to the median) and find the median of these numbers, that is called the upper quartile or quartile 3.
- The interquartile range is the difference between quartile 1 and quartile 3.
- 4, 5, 6, 8, 9, 10, 11, 13
- Median=8.5
- Q1=6.5
- Q3=10.5
- 10.5-6.5=4
I also added my PowerPoint if you didn't get what I was saying.
The are of the purple figure is 18 square units.
Step-by-step explanation:
Just count the number of squares of the purple units. Flr the incomplete squares, find the corresponding incomplete square and count them as 1.
Answer:
-50x - 66
Step-by-step explanation:
Answer:
Null Hypothesis: H_0: \mu_A =\mu _B or \mu_A -\mu _B=0
Alternate Hypothesis: H_1: \mu_A >\mu _B or \mu_A -\mu _B>0
Here to test Fertilizer A height is greater than Fertilizer B
Two Sample T Test:
t=\frac{X_1-X_2}{\sqrt{S_p^2(1/n_1+1/n_2)}}
Where S_p^2=\frac{(n_1-1)S_1^2+(n_2-1)S_2^2}{n_1+n_2-2}
S_p^2=\frac{(14)0.25^2+(12)0.2^2}{15+13-2}= 0.0521154
t=\frac{12.92-12.63}{\sqrt{0.0521154(1/15+1/13)}}= 3.3524
P value for Test Statistic of P(3.3524,26) = 0.0012
df = n1+n2-2 = 26
Critical value of P : t_{0.025,26}=2.05553
We can conclude that Test statistic is significant. Sufficient evidence to prove that we can Reject Null hypothesis and can say Fertilizer A is greater than Fertilizer B.
