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ikadub [295]
4 years ago
8

Calculate the difference between the 1990-1997 average blue crab population and the 1998-2009 average blue crab population predi

ct the average blue crab population for 2010-2020 and explain
Mathematics
1 answer:
Karolina [17]4 years ago
5 0
The difference of the blue crab population From 1990-1997, the average was (800+750)/2 = 775; and from 1989-2009: (350+350)/2 = 350. I predict that from 2010 to 2020 the population will increase to about 500 because it seems to go on trends of increasing and decreasing.
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Antwon makes hot chocolate by mixing 2 cups of milk with 5 tablespoons of cocoa.
Yuki888 [10]

Answer:

2.5 tablespoons of cocoa

Step-by-step explanation:

if 2 cups of milk is originally used then you use 1 well 2 divided by 2 equals 1 and 5 divided by 2 equals 2.5 so you would use 2.5 tablespoons of cocoa

6 0
3 years ago
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Read the paragraphs.
GuDViN [60]

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Uninsured people are less likely to receive care and more likely to have poor health status.

Step-by-step explanation:

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3 years ago
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Determine whether a probability distribution is given. If a probability distribution is given, find its mean and standard deviat
drek231 [11]

Answer:

E(X) = \sum_{i=1}^n X_i P(X_i) = 0*0.031 +1*0.156+ 2*0.313+3*0.313+ 4*0.156+ 5*0.031 = 2.5

We can find the second moment given by:

E(X^2) = \sum_{i=1}^n X^2_i P(X_i) = 0^2*0.031 +1^2*0.156+ 2^2*0.313+3^2*0.313+ 4^2*0.156+ 5^2*0.031 =7.496

And we can calculate the variance with this formula:

Var(X) =E(X^2) -[E(X)]^2 = 7.496 -(2.5)^2 = 1.246

And the deviation is:

Sd(X) = \sqrt{1.246}= 1.116

Step-by-step explanation:

For this case we have the following probability distribution given:

X          0            1        2         3        4         5

P(X)   0.031   0.156  0.313  0.313  0.156  0.031

The expected value of a random variable X is the n-th moment about zero of a probability density function f(x) if X is continuous, or the weighted average for a discrete probability distribution, if X is discrete.

The variance of a random variable X represent the spread of the possible values of the variable. The variance of X is written as Var(X).  

We can verify that:

\sum_{i=1}^n P(X_i) = 1

And P(X_i) \geq 0, \forall x_i

So then we have a probability distribution

We can calculate the expected value with the following formula:

E(X) = \sum_{i=1}^n X_i P(X_i) = 0*0.031 +1*0.156+ 2*0.313+3*0.313+ 4*0.156+ 5*0.031 = 2.5

We can find the second moment given by:

E(X^2) = \sum_{i=1}^n X^2_i P(X_i) = 0^2*0.031 +1^2*0.156+ 2^2*0.313+3^2*0.313+ 4^2*0.156+ 5^2*0.031 =7.496

And we can calculate the variance with this formula:

Var(X) =E(X^2) -[E(X)]^2 = 7.496 -(2.5)^2 = 1.246

And the deviation is:

Sd(X) = \sqrt{1.246}= 1.116

6 0
3 years ago
Hans deposits $300 into an account that pays simple interest at a rate of 2% per year. How much interest will he be paid in the
blagie [28]
300 x 0.02 x 5 = 30
answer
<span> interest will $30 in the first 5 years</span>
5 0
3 years ago
Will spent $3.75 for 3 pounds of granola. What is his unit rate in dollars per pound?
Drupady [299]
The unit rate is $1.25 per pound
6 0
3 years ago
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