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Answer:
3.600
Step-by-step explanation:
0.25(x+1.6)+5x=18
Open the bracket first
0.25x+0.72+5x=18
Collect like terms
0.25x+5x=18-0.72
5.25x=18.72
Divide both sides by 5.25
x=3.566~3.600(nearest hundredth)
Answer:
177
Step-by-step explanation:
This scenario can be modeled as an <u>exponential function</u>.
General form of an exponential function: 
where:
- a is the initial value (y-intercept)
- b is the base (growth/decay factor) in decimal form
- x is the independent variable
- y is the dependent variable
If b > 1 then it is an increasing function
If 0 < b < 1 then it is a decreasing function
If the number of trees <u>increase</u> by <u>10% each year</u>, then the number of trees each year will be 110% of the number of trees the previous year. Therefore, the <u>growth factor is 110%</u>.
Given:
- a = 100 trees
- b = 110% = 1.10 (in decimal form)
- x = time (in years)
- y = number of trees in the orchard
Substituting the given values into the function:

(where x is time in years and y is the number of trees in the orchard)
To find how many trees are in the orchard in the 6th year, input x = 6 into the found equation:

Therefore, Martin had 177 trees in his orchard in the sixth year.
Answer:
0.5328 = 53.28% probability that the length of hatching times is between 15 and 18 days.
Step-by-step explanation:
Normal Probability Distribution:
Problems of normal distributions can be solved using the z-score formula.
In a set with mean
and standard deviation
, the z-score of a measure X is given by:

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.
Mean of 16 days and standard deviation of 2 days.
This means that 
Probability that the length of hatching times is between 15 and 18 days.
This is the p-value of Z when X = 18 subtracted by the p-value of Z when X = 15. So
X = 18



has a p-value of 0.8413
X = 15



has a p-value of 0.3085
0.8413 - 0.3085 = 0.5328
0.5328 = 53.28% probability that the length of hatching times is between 15 and 18 days.