Complete Question
The weights of soy patties sold by a diner are normally distributed. A random sample of 15 patties yields a mean weight of 3.8 ounces with a sample standard deviation of 0.5 ounces. At the 0.05 level of significance,perform a hypothesis test to see if the true mean weight is less than 4 ounce.
Answer:
Yes the true mean weight is less than 4 ounce
Explanation:
From the question we are told that
The random sample is 
The mean weight is 
The standard deviation is 
The level of significance is 
So
The null hypothesis is 
The alternative hypothesis is 
Generally the critical value which a bench mark to ascertain whether the null hypothesis is true or false is mathematically represented as
This value is obtained from the critical value table
Generally the test statistics is mathematically represented as
=> 
So since ST is less than 
then the null hypothesis would be rejected and the alternative hypothesis would be accepted so
Thus the true mean weight is less than 4
To develop this problem we will apply the concepts related to Sound Intensity. This can be defined as the power carried by sound waves per unit area in a direction perpendicular to that area. Mathematically it can be written as,
= Level Difference
n = Number of equal loud sound sources
n = 15
Therefore replacing we have,
Therefore over 100dB the total difference intensity level would be
Here,
The intensity level generated by one engine is
Answer: Sea otter is the pioneer species in the kelp forest as it regulates and controls the population of other species in the kelp forest.
Explanation:
If sea otters are hunted and their population is brought to extinction then this will cause major harm the ecosystem of the kelp forest and it will disturb the ecological balance in the kelp forest. The herbivorous animals consumed by the sea otters will increase in population and they will consume a lot of vegetation in the forest. The kelp forest which forms the coastline will not remain effective in providing protection against the storms to the neighboring areas.
The correct answer is c) 28 m/s.
Let's find the step-by-step solution. The motion of the monkey is an uniformly accelerated motion, with acceleration equal to
. The initial velocity of the monkey is zero, while the distance covered is S=40 m. Therefore, we can use the following relationship to find vf, the final velocity of the monkey:
from which
