Answer :

Step-by-step explanation :
To find the product of

First we expand the bracket ,
it implies that, we use the expression outside the bracket to multiply individual expressions inside the bracket.
Hence


we now apply the law of indices

meaning, when you are multiplying two expressions with the same bases , repeat one of the bases and add the exponents.
Then, simplify to obtain
Thats easy. figure it out
Answer:
a) Statistic.
b) The population proportion is expected to be between 0.29 and 0.31 with a 94% degree of confidence.
Step-by-step explanation:
a) The proportion of 30% is a statistic, as it is a value that summarizes data only from the sample taken in the study from USA Today. Other samples may yield different proportions.
b) We can use the statistic to estimate a confidence interval for the parameter of the population.
The standard error for the proportion is calculated as:

The margin of error is 0.01. We can use this value to determine the level of confidence that represents.
The formula for the margin of error is:

This z-value, according to the the standard normal distribution, corresponds to a confidence interval of 94%.
The interval for this margin of error is:

Then, we can conclude that the population proportion is expected to be between 0.29 and 0.31 with a 94% degree of confidence.
The answer is 56.4 to the nearest tenth
Answer:
£6000
Step-by-step explanation:
Given:
Phillip Cage played, £21,000 was split at a ratio of 2:5 between the arena and Phillip Cage.
Question asked:
How much money did the arena make from Phillip Cage's concert?
Solution:
Ratio Arena and Philip cage in which money distributed = 2 : 5
Let ratio be 
Money earned by Arena = 
Money earned by Phillip Cage = 
Phillip Cage played total for = £21,000
<u>Money earned by Arena + Money earned by Phillip Cage = £21,000</u>

Dividing both sides by 7

Money earned by Arena =
= 
Thus, Arena made £6000 from Phillip Cage's concert.