Answer:
It is very unusual as the probability to get a sample average of 75 or more customers if the manager had not offered the discount is 0.006
Step-by-step explanation:
We are given the following information in the question:
Mean, μ = 50
Standard Deviation, σ = 10
We are given that the distribution of number of customers is a bell shaped distribution that is a normal distribution.
In an attempt to increase the number of weekday customers, the manager offers a $10 membership discount on 5 consecutive weekdays and the sample mean of customers during this weekday period jumps to 75.
Formula:
a) P(sample average of 75 or more customers if the manager had not offered the discount)
P(x > 75)
Calculation the value from standard normal z table, we have,
It is very unusual as the probability to get a sample average of 75 or more customers if the manager had not offered the discount is 0.006
b) Yes, the managers discount strategy worked as it increased the average number of customers from 50 customers to 75 customers.
Sure! Let's do this in order.
First, we can simplify the left side with the distributive property, getting us
8c-14
Next, we can simplify the right side with the distributive property, getting us
8c-8-6.
That gives us 8c-14<8c-8-6.
We can subtract 8c from both sides, getting us
-14<-8-6
Simplify the right side to get
-14<-14
This inequality has no solution. This is because the variable is gone, and the remaining equation isn't true.
300 carrots. You want to find the area of the garden, which you can find by adding the area of the shapes within the garden. You can split the garden into two triangles and a square. The two triangles are equal and form a square with sides of 2 square units, for an area of 4 square units. The square has a side of 4 square units, for an area of 16 square units. You add the 4 square units and the 16 square units to get a total of 20 square units. You would then multiply the 20 square units by the 15 carrots per square unit to get a total of 300 carrots in the garden.
This question requires us to use the cosine rule:
a^2 = b^2 + c^2 - 2bc*cos(A),
where A is the included angle between sides b and c, and a is the side of the triangle opposite to the angle.
In the context of the question, a is the length of the tunnel (let's call this t), b is 6 km, c is 7 km and A is 29°.
Given the values in the question and those we defined, we can rewrite the equation for the cosine rule as:
t^2 = 6^2 + 7^2 - 2(6)(7)cos(29)
Now, evaluating this we get:
t^2 = 36 + 49 - 84cos(29)
t^2 = 85 - 84cos(29)
t = sq.root (85 - 84cos(29))
= 3.40 km (rounded to two decimal places)
