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Complete Question
The lifespans of lions in a particular zoo are normally distributed. The average lion lives 12.5 years; the standard deviation is 2.4 years. Use the empirical rule (68-95-99.7\%)(68−95−99.7%) to estimate the probability of a lion living between 5.3 to 10. 1 years.
Answer:
Thehe probability of a lion living between 5.3 to 10. 1 years is 0.1585
Step-by-step explanation:
The empirical rule formula states that:
1) 68% of data falls within 1 standard deviation from the mean - that means between μ - σ and μ + σ.
2) 95% of data falls within 2 standard deviations from the mean - between μ – 2σ and μ + 2σ.
3) 99.7% of data falls within 3 standard deviations from the mean - between μ - 3σ and μ + 3σ.
Mean is given in the question as: 12.5
Standard deviation : 2.4 years
We start by applying the first rule
1) 68% of data falls within 1 standard deviation from the mean - that means between μ - σ and μ + σ.
μ - σ
12.5 -2.4
= 10.1
We apply the second rule
2) 95% of data falls within 2 standard deviations from the mean - between μ – 2σ and μ + 2σ.
μ – 2σ
12.5 - 2 × 2.4
12.5 - 4.8
= 7.7
We apply the third rule
3)99.7% of data falls within 3 standard deviations from the mean - between μ - 3σ and μ + 3σ.
μ - 3σ
= 12.5 - 3(2.4)
= 12.5 - 7.2
= 5.3
From the above calculation , we can see that
5.3 years corresponds to one side of 99.7%
Hence,
100 - 99.7%/2 = 0.3%/2
= 0.15%
And 10.1 years corresponds to one side of 68%
Hence
100 - 68%/2 = 32%/2 = 16%
So,the percentage of a lion living between 5.3 to 10. 1 years is calculated as 16% - 0.15%
= 15.85%
Therefore, the probability of a lion living between 5.3 to 10. 1 years
is converted to decimal =
= 15.85/ 100
= 0.1585
Answer:
Part A) 
Part B) She will spend more than 
Step-by-step explanation:
Let
p-----> the number of hamburger patties
Part A) Luna has to buy at least 16 packages for an upcoming picnic
Part B) Suppose she actually needs more than 150 hamburgers. How much will she spend?
Let
c---------> the total cost
step 1
Divide 150 hamburgers by 8 (a package of hamburgers)
so
round to the nearest whole number
----> the minimum number of packages
step 2
1) The outcomes for rolling two dice, the sample space, is as follows:
(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6)
(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6)
(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6)
(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6)
(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)
(6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)
There are 36 outcomes in the sample space.
2) The ways to roll an odd sum when rolling two dice are:
(1, 2), (1, 4), (1, 6), (2, 1), (2, 3), (2, 5), (3, 2), (3, 4), (3, 6), (4, 1), (4, 3), (4, 5), (5, 2), (5, 4), (5, 6), (6, 1), (6, 3), (6, 5). There are 18 outcomes in this event.
3) The probability of rolling an odd sum is 18/36 = 1/2 = 0.5
In geometry, similar figures are those whose ratios of the corresponding sides are equal and the corresponding angles are congruent. In relation to the volume, we determine first the cube roots of the given and find the ratio as shown below.
s1 / s2 = cube root of (512/343)
= 8/7
The square of this ratio is the ratio of the areas of the figure. If we let x be the area of the smaller figure then,
(8/7)^2 = 192 mm²/ x
The value of x from the equation is 147 mm².
The area therefore of the smaller figure is 147 mm².
