t = number of tickets
15t ≥ 1500
so....
step 1. 15t ≥ 1500 --> divide both sides by 15 --> 15t/15 ≥ 1500/15
step 2. t ≥ 100
So 100 at the least, ticket should be sold to reach atleast $1500.
I hope this helps. :)
Answer:
a) Null Hypothesis: length of each screw is less than 14 centimeters
b) Alternate Hypothesis: length of each screw is equal to or greater than 14 centimeters
Step-by-step explanation:
Complete question
A factory that manufactures screws is performing a quality control experiment. Each object should have a length of no more than 14 centimeters. The factory believes that the length of the screws exceeds this value and measures the length of screws. The sample mean screw length was centimeters. The population standard deviation is known to be centimeters.
1. What is the null hypothesis?
2. What is the alternative hypothesis?
Solution :
The null hypothesis is basically the problem statement that needs to be tested.
Alternate hypothesis is opposite of that of null hypothesis
a) Null Hypothesis: length of each screw is less than 14 centimeters
b) Alternate Hypothesis: length of each screw is equal to or greater than 14 centimeters
Answer:
Honestly I don’t know I need help
Step-by-step explanation:
SOLUTION
From the given data the mean is 62 and standard deviation is 4
It is required to find the probability that a data value is between 57 and 62
That is:

The z scores is calculated using:

Using the x values it follows:

Also,

Thus the required probability is:
![P(-1.25The proability is:[tex]P(-1.25This can be expressed as percentage as:[tex]P(-1.25\lt z\lt0.75)=66.8\%](https://tex.z-dn.net/?f=P%28-1.25The%20proability%20is%3A%5Btex%5DP%28-1.25This%20can%20be%20expressed%20as%20percentage%20as%3A%5Btex%5DP%28-1.25%5Clt%20z%5Clt0.75%29%3D66.8%5C%25)
Therefore the correct option is C