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maxonik [38]
3 years ago
11

A researcher is studying a mouse population in a field. For the first year of her study, she notices that the mouse population d

emonstrates exponential growth. However, by the third year of her study, the mouse population has stabilized at approximately 500 mice (although this number fluctuates a bit). Thus, 500 mice would be the _____ of the field being studied.
Mathematics
1 answer:
Montano1993 [528]3 years ago
5 0

Answer:

The 500 mice would be the carrying capacity of the field being studied.

Step-by-step explanation:

For the first year of study, the mouse population demonstrates exponential growth.

But by third year, the mouse population has stabilized at approximately 500 mice.

Thus, this 500 mice would be the carrying capacity of the field being studied.

Carrying capacity is defined as the measure to demonstrate the maximum population size that any given environment can sustain for an unspecified time period.

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Claire is buying a new bicycle for 295 if the sales tax is 4.75% what will she pay in this total
yawa3891 [41]

Answer:

the answer 14.0125. rounded = 14

Step-by-step explanation:

convert the percentage to a decimal

4.75% --> move two decimal places --> 0.0475

multiply 295*0.0475

= 14.0125

6 0
3 years ago
Factor the expression using the GCF.<br> 60 – 36 =
Ivanshal [37]

Answer

12 (5 - 3)

Step-by-step explanation:

The greatest common factor is the largest number that will divide both numbers evenly. As an example, 6 is a common factor because it divides both 60 and 36. However, it is not the greatest common factor.

 

If I break 60 into it's prime factors, I would get 2 * 2 * 3 * 5.

 

If I break 36 into it's prime factors, I would get 2 * 2 * 3 * 3.

 

The two number have the factors, 2, 2, and 3 in common.

 

The greatest common factor is then: 2 * 2 * 3, or 12.

 

The expression can be factored as:  12 ( 5 - 3)

6 0
3 years ago
During a building project, a brick is
taurus [48]

The brick will fall to the ground after 9 secs.

<h3>What are quadratic equations?</h3>

Given the equation for height as a function of time is  h(t) = -16t2 + 1,296

The height of the brick on the ground is h(t) = 0. Substitute h(t) = 0 into the equation as shown:

  • -16t2 + 1,296 = 0

Subtract 1296 from both sides

-16t² + 1,296 - 1296 = 0 - 1296

16t² = 1296

t² = 1296/16

t² = 81

t = 9secs

Hence the brick will fall to the ground after 9 secs.

Learn more on equations here: brainly.com/question/13763238

3 0
2 years ago
13x + 7x - 4x = (-40)
dlinn [17]
13x + 7x - 4x = -40
combine like terms o0n the left
16x = -40
divide both sides by 16
x = -5/2 or -2.5
5 0
3 years ago
Read 2 more answers
A random sample of 49 measurements from one population had a sample mean of 15, with sample standard deviation 5. An independent
kotykmax [81]

Answer:

Comparing the p value with the significance level \alpha=0.01 we see that p_v so we can conclude that we have enough evidence to reject the null hypothesis, and the difference between the two groups is significantly different.  

Step-by-step explanation:

\bar X_{1}=15 represent the mean for sample 1

\bar X_{2}=18 represent the mean for sample 2

s_{1}=5 represent the sample standard deviation for 1  

s_{f}=6 represent the sample standard deviation for 2  

n_{1}=49 sample size for the group 2  

n_{2}=64 sample size for the group 2  

\alpha=0.01 Significance level provided

t would represent the statistic (variable of interest)  

Concepts and formulas to use  

We need to conduct a hypothesis in order to check if the difference in the population means, the system of hypothesis would be:  

Null hypothesis:\mu_{1}-\mu_{2}=0  

Alternative hypothesis:\mu_{1} - \mu_{2}\neq 0  

We don't have the population standard deviation's, so for this case is better apply a t test to compare means, and the statistic is given by:  

t=\frac{(\bar X_{1}-\bar X_{2})-\Delta}{\sqrt{\frac{s^2_{1}}{n_{1}}+\frac{s^2_{2}}{n_{2}}}} (1)

And the degrees of freedom are given by df=n_1 +n_2 -2=49+64-2=111  

t-test: Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other.

With the info given we can replace in formula (1) like this:  

t=\frac{(15-18)-0}{\sqrt{\frac{5^2}{49}+\frac{6^2}{64}}}}=-2.897

P value

Since is a bilateral test the p value would be:  

p_v =2*P(t_{111}  

Comparing the p value with the significance level \alpha=0.01 we see that p_v so we can conclude that we have enough evidence to reject the null hypothesis, and the difference between the two groups is significantly different.  

7 0
3 years ago
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