Divide using long division. (-2x^3-4x^2-4x+2) divide by (x-4)
1 answer:
You might be interested in
THE ANSWER IS $22.01
Answer:
The probability of such a tire lasting more than 60,000 miles is 0.0228, for the complete question provided in explanation.
Step-by-step explanation:
<h2>Q. This is the question: </h2>The lifetime of a certain type of car tire are normally distributed. The mean lifetime of a car tire is 50,000 miles with a standard deviation of 5,000 miles. Assume that the manufacturer's claim is true, what is the probability of such a tire lasting more than 60,000 miles?
<h2>Answer:</h2>This is the question of normal distribution:
First w calculate the value of Z corresponding to X = 60,000 miles
We, have; Mean = μ = 50,000 miles, and Standard Deviation = σ = 5,000 miles
Now, for Z, we know that:
Z = (x-μ)/σ
Z = (60,000 - 50,000)/5,000
<u>Z = 2</u>
Now, we have standard tables, for normal distribution in terms of values of Z. One is attached in this answer.
P(X > 60,000) = P(Z > 2) = 1 - P(Z = 2)
P(X > 60,000) = 1 - 0.9772
<u>P(X > 60,000) = 0.0228</u>
Answer:
Half-life of the goo is 49.5 minutes
191.7 grams of goo will remain after 32 minutes
Step-by-step explanation:
Let denotes initial and final mass.
According to exponential decay,
Here, t denotes time and k denotes decay constant.
So, half-life of the goo in minutes is calculated as follows:
Half-life of the goo is 49.5 minutes
So,
Put
Put t = 32 minutes