1. a
2. c
3. b
hope this helps
<u>The television station has received/A many complaints about/B the clothing advertisements, which some/C viewers condemn to be/D </u><u>tasteless</u><u>. No error/E</u>
The correct option is D.
- The television station has received many complaints about the clothing advertisements, which some viewers complained were tasteless.
- Well, you could mean that Writing Question is a 'trick one' .
What is a television station called?
- Most often the term "television station" refers to a station which broadcasts structured content to an audience or it refers to the organization that operates the station.
- A terrestrial television transmission can occur via analog television signals or, more recently, via digital television signals.
Learn more about television station
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Answer:
Step-by-step explanation:
42 < 21+3x < 84
21 < 3x < 63
7 < x < 21
Answer: Average rates of change for Bird A is 0.045 and for Bird B is 3.816.
Step-by-step explanation: Through its graph, a function can be analyzed and be identified its attributes. Average Rate of Change is the ratio of change in the function values to the change of x-value, i.e., it is the slope of the function in a specific interval of x. With the Average it is possible to compare two function.
<u>Average Rate of Change</u>
<em><u>Bird A</u></em>:
For the interval [0,18]:
x₁ = 0 f(0) = 8.3
x₂ = 18 f(18) = 9.1
Average = 
Average = 
Average = 0.045
<em><u>Bird B</u></em>: y = 3.6(1.06)x
For the interval [0,18]:
x₁ = 0 y = 3.6(1.06)0 = 0
x₂ = 18 y = 3.6(1.06)18 = 68.7
Average =
Average = 
Average = 3.816
The Average Rate for Change for Bird A is 0.045 and for Bird B is 3.816. This means that the population of Bird B increase in rate faster than the population of Bird A.
Answer:
Step-by-step explanation:
Hello!
X: the lifespan of a new computer monitor of Glotech.
The average life is μ= 85 months and the variance δ²= 64
And a sample of 122 monitors was taken.
You need to calculate the probability that the sample mean is greater than 86.6 months.
Assuming that the variable has a normal distribution X~N(μ;δ²), then the distribution of the sample mean is X[bar]~N(μ;δ²/n)
To calculate this probability you have to work using the sampling distribution and the following formula Z= (X[bar]-μ)/δ/√n ~N(0;1)
P(X[bar]>86.6)= 1 - P(X[bar]≤86.6)
1 - P(Z≤(86.6-85)/(8/√122))= 1 - P(Z≤2.21)= 1 - 0.98645= 0.013355
The probability of the sample mean is greater than 0.013355
I hope this helps!
