Answer: x = 1/16
Step-by-step explanation:
Since the inverse of a Logarithm is an exponential function, we know that the final solution has to involve an exponential function somewhere in it.
1. log B(2) {x} = -4 || given
2. x = 2 ^ -4 || Logarithm rule that allows you to move the base of the logarithm to the base of the exponent on the other side. For example, if you had log B(5) {x} = 3, the base of 5 would move over to the other side and it would be raised to 3; x = 5^3.
3. x = (1) / (2^4) || Simplify. Use the negative exponent rule. This rule always leaves a numerator of 1, and a denominator of your exponent. In this case, it will be 2 ^ -4, so you will do 2^4 which is 16 and you will put that over 1. Resulting in your final answer of x = 1/16
<span><span><span><span>2x</span>+y</span>+<span>−y</span></span>=<span>6+<span>−y</span></span></span><span><span>2x</span>=<span><span>−y</span>+6</span></span>Step 2: Divide both sides by 2.<span><span><span>2x</span>2</span>=<span><span><span>−y</span>+6</span>2</span></span><span>x=<span><span><span><span>−1</span>2</span>y</span>+3</span></span>Answer:<span> x=<span><span><span><span>−1</span>2</span>y</span>+3</span></span>
x-3=-2
x-3+3=-2+3
answer x=1
<span>(a) This is a binomial
experiment since there are only two possible results for each data point: a flight is either on time (p = 80% = 0.8) or late (q = 1 - p = 1 - 0.8 = 0.2).
(b) Using the formula:</span><span>
P(r out of n) = (nCr)(p^r)(q^(n-r)), where n = 10 flights, r = the number of flights that arrive on time:
P(7/10) = (10C7)(0.8)^7 (0.2)^(10 - 7) = 0.2013
Therefore, there is a 0.2013 chance that exactly 7 of 10 flights will arrive on time.
(c) Fewer
than 7 flights are on time means that we must add up the probabilities for P(0/10) up to P(6/10).
Following the same formula (this can be done using a summation on a calculator, or using Excel, to make things faster):
P(0/10) + P(1/10) + ... + P(6/10) = 0.1209
This means that there is a 0.1209 chance that less than 7 flights will be on time.
(d) The probability that at least 7 flights are on time is the exact opposite of part (c), where less than 7 flights are on time. So instead of calculating each formula from scratch, we can simply subtract the answer in part (c) from 1.
1 - 0.1209 = 0.8791.
So there is a 0.8791 chance that at least 7 flights arrive on time.
(e) For this, we must add up P(5/10) + P(6/10) + P(7/10), which gives us
0.0264 + 0.0881 + 0.2013 = 0.3158, so the probability that between 5 to 7 flights arrive on time is 0.3158.
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Answer:
If you go to a website with a graph place the points and look where the line sets
Step-by-step explanation:
Answer:
NOt finished
Step-by-step explanation:
PAGE 1
1.) enlargement b/c the number is more that 1
2.) reduction b/c the number is less than 1
3.) enlargement b/c the number is more than 1
4.) reduction b/c the number is less than 1
5.) reduction k = 1/3
6.) enlargement k = 5/2 or 2.5
PAGE 2
7.) reduction k = 1/3
8.) enlargement k = 2
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I WILL DO THE SCREENSHOTS TOMORROW
