Answer:
C=(x – 1)(x2 + x + 1)
E=(x + 4)(x2 – 4x + 16)
Answer:
D
Step-by-step explanation:
It's d because it has the highest x value. The higher x value is the farther it is from the y-axis
Basically that answer is incorrect. If you'd learn about combining, multiplying, dividing or subtracting like terms, this will be a simple problem. To be specific, like term is a number with a letter with it.
For example, 2.4z is a like term.
so you can only add like term, not any other normal number. When two terms have the same "letter", you will have to do what the sign said so in the problem....
2.4z + 1.2z - 6.5 = 0.7
3.6z - 6.5 = 0.7 (Add but leave the z alone because you are not multiplying)
The final answer is 3.6z - 6.5 because you cannot do anything else to that problem.
3.6z-6.5=0.7 is not true.
Answer:
1 1/4 is your answer.
Step-by-step explanation:
What you do is you divide.
4 goes into 5 1 time with 1 left over.
1 1/4 is your answer.
A) maximum mean weight of passengers = <span>load limit ÷ number of passengers
</span><span>
maximum mean weight of passengers = 3750 </span>÷ 25 = <span>150lb
</span>B) First, find the z-score:
z = (value - mean) / stdev
= (150 - 199) / 41
= -1.20
We need to find P(z > -1.20) = 1 - P(z < -1.20)
Now, look at a standard normal table to find <span>P(z < -1.20) = 0.11507, therefore:
</span>P(z > -1.20) = 1 - <span>0.11507 = 0.8849
Hence, <span>the probability that the mean weight of 25 randomly selected skiers exceeds 150lb is about 88.5%</span> </span>
C) With only 20 passengers, the new maximum mean weight of passengers = 3750 ÷ 20 = <span>187.5lb
Let's repeat the steps of point B)
z = (187.5 - 199) / 41
= -0.29
P(z > -0.29) = 1 - P(z < -0.29) = 1 - 0.3859 = 0.6141
</span>Hence, <span>the probability that the mean weight of 20 randomly selected skiers exceeds 187.5lb is about 61.4%
D) The mean weight of skiers is 199lb, therefore:
number</span> of passengers = <span>load limit ÷ <span>mean weight of passengers
= 3750 </span></span><span>÷ 199
= 18.8
The new capacity of 20 skiers is safer than 25 skiers, but we cannot consider it safe enough, since the maximum capacity should be of 18 skiers.</span>
