Ask question

Ask question

All categories

2 years ago

10

The scatter plot shows the maximum noise level when different numbers of the people are in the stadium. The linear model is give

n by the equation y=1.5x+22.7, where y represents maximum noise level and x represents the number of people the number of people ,in thousands, in the stadium. y=1.5x+22.7
3. What is the y intercept of the linear model given? What does it mean in the context of the problem? Is this reasonable? Explain the reasoning?

1 answer:

Sholpan [36]2 years ago

8 0

The y intercept is 1.5x+22.7 + the linear equation which is 22

You might be interested in

Arada [10]

Answer:

.

Step-by-step explanation:

7 0

2 years ago

Each morning librarians place 560 books on a book shelf in the school library. Each student that visits library that day takes e

Alex73 [517]

Answer:

Step-by-step explanation:

560=3s+b

b.

560=3×150+b

b=560-450=110

when s=149,b=560-3×149=560-447=113

s=2,b=560-3×2=560-6=554

s=1,b=560-3×1=557

s=0,b=560-3×o=560

so domain of b={110,113,116,...,554,557,560}

7 0

3 years ago

The price of grapes, g, is $2.19 per pound. The price of bananas, b, is $0.59 per pound. The price of pears, p, is $1.49 per pou

AlekseyPX

Part A.
Before you can write any sort of expression, you need to define variables. "grapes g" is not a definition, so the exercise seems meaningless as written. It seems the intent is to ...
let g, b, p represent the numbers of pounds of grapes, bananas, and pears, respectively.

Then, the total cost of some weight of fruit is
2.19g + 0.59b + 1.49p

Part B.
For g=3, b=3, p=2, the expression evaluates to
2.19*3 +0.59*3 +1.49*2 = 11.32

The total cost of 3 pounds of grapes, 3 pounds of bananas, and 2 pounds of pears is ...
$11.32

3 0

3 years ago

Which number is greater or equal or less than

Anna71 [15]

We need more to answer the question I think

3 0

3 years ago

A given field mouse population satisfies the differential equation dp dt = 0.5p − 410 where p is the number of mice and t is the

ohaa [14]

Answer:

a) $t = 2 *ln(\frac{82}{5}) =5.595$

b) $t = 2 *ln(-\frac{820}{p_0 -820})$

c) $p_0 = 820-\frac{820}{e^6}$

Step-by-step explanation:

For this case we have the following differential equation:

$\frac{dp}{dt}=\frac{1}{2} (p-820)$

And if we rewrite the expression we got:

$\frac{dp}{p-820}= \frac{1}{2} dt$

If we integrate both sides we have:

$ln|P-820|= \frac{1}{2}t +c$

Using exponential on both sides we got:

$P= 820 + P_o e^{1/2t}$

Part a

For this case we know that p(0) = 770 so we have this:

$770 = 820 + P_o e^0$

$P_o = -50$

So then our model would be given by:

$P(t) = -50e^{1/2t} +820$

And if we want to find at which time the population would be extinct we have:

$0=-50 e^{1/2 t} +820$

$\frac{820}{50} = e^{1/2 t}$

Using natural log on both sides we got:

$ln(\frac{82}{5}) = \frac{1}{2}t$

And solving for t we got:

$t = 2 *ln(\frac{82}{5}) =5.595$

Part b

For this case we know that p(0) = p0 so we have this:

$p_0 = 820 + P_o e^0$

$P_o = p_0 -820$

So then our model would be given by:

$P(t) = (p_o -820)e^{1/2t} +820$

And if we want to find at which time the population would be extinct we have:

$0=(p_o -820)e^{1/2 t} +820$

$-\frac{820}{p_0 -820} = e^{1/2 t}$

Using natural log on both sides we got:

$ln(-\frac{820}{p_0 -820}) = \frac{1}{2}t$

And solving for t we got:

$t = 2 *ln(-\frac{820}{p_0 -820})$

Part c

For this case we want to find the initial population if we know that the population become extinct in 1 year = 12 months. Using the equation founded on part b we got:

$12 = 2 *ln(\frac{820}{820-p_0})$

$6 = ln (\frac{820}{820-p_0})$

Using exponentials we got:

$e^6 = \frac{820}{820-p_0}$

$(820-p_0) e^6 = 820$

$820-p_0 = \frac{820}{e^6}$

$p_0 = 820-\frac{820}{e^6}$

8 0

3 years ago