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valkas [14]
3 years ago
7

Simplify the expression. -1 + (-3) - (-5) =

Mathematics
2 answers:
mestny [16]3 years ago
6 0

Answer:

1

Step-by-step explanation:

-1 + (-3) - (-5)= -1 + 2 =  1  

  - and - give you +, so positive 5 and negative 3

- 3 + 5= 2  

Mashcka [7]3 years ago
5 0

Answer:

=-1)+(-3)-(-5)

=-1-3+5

=-4+5

=1

Step-by-step explanation:

answer is to your question is 1.

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Isle Royale, an island in Lake Superior, has provided an important study site of wolves and their prey. Of special interest is t
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Answer:

<em>Part a) Probability that a moose in that age group is killed by a wolf</em>

  • P  (0.5 year) = 0.361
  • P (1 - 5) = 0.159
  • P (6 - 10) = 0.267
  • P (11 - 15) = 0.203
  • P (16 - 20) = 0.010

<em>Part b)</em>

  • <em>Expected age of a moose killed by a wold</em>

         μ = 5.61 years

  • <em>Stantard deviation of the ages</em>

       σ = 4.97 years

Explanation:

1) Start by arranging the table to interpret the information:

<u>Age of Moose in years          Number killed by woolves</u>

Calf (0.5)                                                 107

1 - 5                                                           47

6 - 10                                                         79

11 - 15                                                        60

16 - 20                                                        3

You can  now verify the total number of moose deaths identified as wolf kills: 107 + 47 + 79 + 60 + 3 = 296, such as stated in the first part of the question.

2) <u>First question: </u>a) For each age, group, compute the probability that a moos in that age group is killed by a wolf.

i) Formula:

Probability = number of positive outcomes / total number of events.

ii) Probability that a moose in an age group is killed by a wolf = number of moose killed by a wolf in that age group / total number of moose deaths identified as wolf kills.

iii) P  (0.5 year) = 107 / 296 = 0.361

iv) P (1 - 5) = 47 / 296 = 0.159

v) P (6 - 10) = 79 / 296 = 0.267

vi) P (11 - 15) = 60 / 296 = 0.203

vii) P (16 - 20) = 3 / 296 = 0.010

3) <u>Second question:</u> b) Consider all ages in a class equal to the class midpoint. Find the expected age of a moose killed by a wolf and the standard deviation of the ages.

i) Class       midpoint

   0.5               0.5

   1 - 5           (1 + 5) / 2 = 3

   6 - 10         (6 + 10) / 2 = 8

   11 - 15         (11 + 15) / 2 = 13

   16 - 20       (16 + 20) = 18

ii) Expected age of a moose killed by a wolf = mean of the distribution = μ

μ = Sum of the products of each probability times its age (mid point)

μ = 0.5 ( 0.361) + 3 ( 0.159) + 8 ( 0.267) + 13 ( 0.203) + 18 ( 0.010) = 5.61 years

μ = 5.61 years ← answer

iii) Stantard deviation of the ages = σ

σ = square root of the variace

variance = s = sum of the products of the squares of the differences between the mean and the class midpoint time the probability.

s =  (0.5 - 5.61)² (0.361) + (3 - 5.61)² ( 0.159) + (8  - 5.61)² ( 0.267) + (13 - 5.61)² ( 0.203) + (18 - 5.61)² ( 0.010) = 24.65

σ = √ (24.65) = 4.97 years ← answer

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A study of pollutants showed that certain industrial emissions should not exceed 2.3 parts per million. You believe a particular
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Answer:

t=\frac{3.3-2.3}{\frac{0.4}{\sqrt{9}}}=7.5    

The degrees of freedom are given by:

df=n-1=9-1=8  

And the p value would be:

p_v =P(t_{(8)}>7.5)=0.0000346  

Since the p value is lower than the significance level we have enough evidence to reject the null hypothesis and we can conclude that the true mean is significantly higher than 2.3 ppm the safe limit

Step-by-step explanation:

Information provided

\bar X=3.3 represent the sample mean in ppm

s=0.4 represent the sample standard deviation

n=9 sample size  

\mu_o =2.3 represent the value to verify

\alpha=0.05 represent the significance level

t would represent the statistic  

p_v represent the p value

System of hypothesis

We want to verify if the true mean is higher than 2,3 ppm, the system of hypothesis would be:  

Null hypothesis:\mu \leq 2.3  

Alternative hypothesis:\mu > 2.3  

The statistic for this case is given by:

t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}  (1)  

Replacing the info given we got:

t=\frac{3.3-2.3}{\frac{0.4}{\sqrt{9}}}=7.5    

The degrees of freedom are given by:

df=n-1=9-1=8  

And the p value would be:

p_v =P(t_{(8)}>7.5)=0.0000346  

Since the p value is lower than the significance level we have enough evidence to reject the null hypothesis and we can conclude that the true mean is significantly higher than 2.3 ppm the safe limit

6 0
3 years ago
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