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nexus9112 [7]
3 years ago
6

Make the equations that model the following situations: A man buys 57 fish to start a fish farm, and that breed of fish is known

to grow it's population by doubling every month.​
Mathematics
1 answer:
sweet [91]3 years ago
3 0
Just square to find them to buy the answer by a dividend then the fraction of the answer then you get your answer
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Please please please help me on this one math problem!! if you dont know it dont answer!
cricket20 [7]

Answer:

fx=

18/5(2.718282−0.1z)+6

mark him as brainlesit

Step-by-step explanation:

i think its right

7 0
3 years ago
Read 2 more answers
Find the slope of the line with x intercept 9 and y intercept of -2
elena-s [515]

Answer:

2/9

Step-by-step explanation:

Find the slope of the line with x intercept 9 and y intercept of -2

Given that the equation of the line is y = mx+b

x intercept occurs when y = 0

The coordinate of x intercept is (9,0)

y intercept occurs at x = 0

The coordinate of y intercept is (0, -2)

Slope m = y2-y1/x2-x1

m = -2-0/0-9

m = -2/-9

m = 2/9

Hence the required slope is 2/9

8 0
3 years ago
Lucinda wants to build a square sandbox, but she has no way of measuring angles. Which of the following explains how she can mak
Sati [7]
All of the angles should be the same because a square has all equal angles and side lengths.
6 0
3 years ago
A design engineer wants to construct a sample mean chart for controlling the service life of a halogen headlamp his company prod
tekilochka [14]

Answer:

C) 515 hours.

D) 500 hours

c) sample 3

Step-by-step explanation:

1. Sample 2 mean = x2`= ∑x2/n2= 2060/4= 515 hours

Sample Service Life (hours)

1                2              3

495      525            470

500         515           480

505        505            460

<u>500         515             470        </u>

<u>∑2000     2060         1880</u>

x1`= ∑x1/n1= 2000/4= 500 hours

x2`= ∑x2/n2= 2060/4= 515 hours

x3`= ∑x3/n3= 1880/4=  470 hours

2. The mean of the sampling distribution of sample means for whenever service life is in control is 500 hours . It is the given mean in the question and the limits are determined by using  μ ± σ , μ±2 σ  or μ ± 3 σ.

In this question the limits are determined by using  μ ± σ .

3. Upper control limit = UCL = 520 hours

Lower Control Limit= LCL = 480 Hours

Sample 1 mean = x1`= ∑x1/n1= 2000/4= 500 hours

Sample 2 mean = x2`= ∑x2/n2= 2060/4= 515 hours

Sample 3 mean = x3`= ∑x3/n3= 1880/4=  470 hours

This means that the sample mean must lie within the range 480-520 hours but sample 3 has a mean of 470 which is out of the given limit.

We see that the sample 3 mean is lower than the LCL. The other  two means are within the given UCL and LCL.

This can be shown by the diagram.

8 0
2 years ago
Factor completely
antiseptic1488 [7]
The frist one is correct for it 
5 0
3 years ago
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