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lapo4ka [179]
3 years ago
7

Part A: Explain why the x-coordinates of the points where the graphs of the equations y = 8^x and y = 2^x+2 intersect are the so

lutions of the equation 8^x = 2^x+2. (4 points)
Part B: Make tables to find the solution to 8^x = 2^x+2. Take the integer values of x between −3 and 3. (4 points)

Part C: How can you solve the equation 8^x = 2^x+2 graphically? (2 points)
Mathematics
1 answer:
Butoxors [25]3 years ago
6 0
If they y values are the same then you can equate these 2 equations y = 8x and y = 2x + 2 8x=2x+2
because where they inersect their x and y values are the same

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12 3/8 divided by 3/4
oksano4ka [1.4K]

Answer:

32/2 or 16.5

Step-by-step explanation:

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3 years ago
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There is a construction zone on a highway. The speeds of vehicles passing through this construction zone are normally distribute
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The percentage of vehicles passing through this construction zone that are traveling at a speed of 50 and 57 miles per hour
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3 years ago
A manufacturer of processing chips knows that 2\%2%2, percent of its chips are defective in some way. Suppose an inspector rando
kipiarov [429]

The data in the question seems a bit erroneous. I am writing the correct question below:

A manufacturer of processing chips knows that 2%, percent of its chips are defective in some way. Suppose an inspector randomly selects 4 chips for an inspection. Assuming the chips are independent, what is the probability that at least one of the selected chips is defective? Lets break this problem up into smaller pieces to understand the strategy behind solving it.

Answer:

The probability that at least one of the selected chips is defective is 0.0776.

Step-by-step explanation:

The question states that the probability of defective chips is 2% i.e. 0.02. Let p denote the probability of selecting a defective chip so, p = 0.02

An inspector selects 4 chips, which means n=4 and we need to compute the probability that at least one of the selected chips is defective. Let X be the number of defective chips selected. We need to compute P(X≥1) which means either 1, 2, 3 or 4 chips can be defective.

We will use the binomial distribution formula to solve this problem. The formula is:

<u>P(X=x) = ⁿCₓ pˣ qⁿ⁻ˣ</u>

where n = total no. of trials

          p = probability of success

          x = no. of successful trials

          q = probability of failure = 1-p

we have n=4, p=0.02 and q=1-0.02=0.98.

We need to compute P(X≥1) which is equal to:

P(X≥1) = P(X=1) + P(X=2) + P(X=3) + P(X=4)

A shorter method to do this is to use the total probability theorem:

P(X≥1) = 1 - P(X<1)

          = 1 - P(X=0)

          = 1 - ⁴C₀ (0.02)⁰(0.98)⁴⁻⁰

          = 1 - (0.98)⁴

          = 1 - 0.9224

P(X≥1) = 0.0776

4 0
3 years ago
How is the answer 35? How do I get it?
vlabodo [156]
I think it's -35. The steps to get that number is subtracting the 3 to the right side. Then you have -2 - 3 which equals to -5. Then you still have x/7= -5. Get rid of 7 from the x and do the same thing to the other side but you multiply 7 and -5. Last you your would be x= -35.
7 0
3 years ago
Divide the polynomial (x^2 + 3x + 5) by 2x + 1​
umka21 [38]

Answer:

50888888888888888888+++++

7 0
3 years ago
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