Answer:
Step-by-step explanation:
One unit up: y = x + 1
One unit down: y = x - 1
One unit to the left: y = x + 1
One unit to the right: y = x - 1
The answer is 0.49
0.7 x 0.7 = 0.7^2
The Type B trees produces 120 pears.
<h3>What is meant by percentage?</h3>
A percentage is a number or ratio expressed as a fraction of 100 in mathematics. The percent sign, "%," is commonly used, but the abbreviations "pct.", "pct," and sometimes "pc" are also used. A percentage is a number with no dimensions; it has no unit of measurement.
<h3>
How many pears did the Type B trees produce?</h3>
- Type A trees produced 20 percent more pears than Type B trees did
- Type A trees produced 144 pears.
Let us consider, Type B produced 100% pears.
So, Type A will produce 120%of B, that is 144
144 = 120/100 
B
simplifying the above equation, we get
144 100 = 120B
B = 14400/120
B = 120
The value of B = 120.
Hence, the Type B trees produces 120 pears
Therefore, the correct answer is option B) 120.
To learn more about percentage, refer to:
brainly.com/question/24304697
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<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.
</span>
Answer:
No solution
Step-by-step explanation:
