So,
To what percent 34 is of 55, we first need to find the fraction equivalent.
34 out of 55 parts or 34/55
Divide 34 by 55.
34/55 = 0.61818...
Now, we move the decimal point 2 places to the right in order to convert it to a percent.
6.1818...
61.81818...
Rounded to the nearest hundredth: 61.82%
Rounded to the nearest tenth: 61.8%
Rounded to the nearest percent: 62%
Find the amount of the difference and then divide it by the original amount:
380 - 350 = 30
30 / 350 = 0.0857 = 8.6% increase. Round the answer as needed.
Answer:
Step-by-step explanation:
Hello, great question. These types are questions are the beginning steps for learning more advanced Probability Problems.
Based on the study that Shane conducted we can assume a couple of things. Firstly, we can see that a vast majority of shoppers are buying products they saw advertised, so we can assume that advertising a product directly affects and contributes to the sales of that particular product.
Secondly, we can tell from the study that the number 1 method for advertising products is the Internet. Since 70% of all the interviewed shoppers are buying a product they saw on the internet.
I hope this answered your question. If you have any more questions feel free to ask away at Brainly.
Answer:
a rotation about point H
Step-by-step explanation:
i just took the test
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.
