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1 year ago

10

You keep track of the daily hot chocolate sales and the outside temperature each day. The data you gathered is shown in the data

table below.
Hot Chocolate Sales and Outside Temperatures
Sales ($) $100 $213 $830 $679 $209 $189 $1,110 $456 $422 $235 $199
Temperature (°F) 92° 88° 54° 62° 85° 16° 52° 65° 68° 89° 91°
how would you describe the correlation in the data? Explain your reasoning

1 answer:

vazorg [7]1 year ago

6 0

Answer:

There is a negative correlation.

Step-by-step explanation:

The point on the scatter plot are downwards sloping. This implies that there is a negative relationship between the variables temperature and sales. Thus, there is a negative correlation.

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Figure A is translated 3 units right and 2 units up. The translated figure is labeled figure B. Figure B is reflected over the x

mojhsa [17]

Answer:

A Is congruent to B and B Is congruent to C

Step-by-step explanation:

Given Figures A, B and C on a coordinate plane with the coordinates below:

Triangle A has points (1, -2), (3, -2), (3, -5).

Triangle B has points (4, 0), (6, 0), (6, -3).

Triangle C has points (4, 0), (6, 0), (6, 3).

The fact that each triangle is a right triangle only shows that the triangles are similar.

For any two shapes to be congruent, they must have in addition to the same shape, the same size.

Therefore, for Triangle A to be congruent to Triangle C, Triangle A must be congruent to Triangle B and Triangle B must be congruent to Triangle C.

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2 years ago

A polyhedron with one pentagonal base is a

alekssr [168]

I think it is an apex

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3 years ago

Square root of -25 in imaginary form

Anvisha [2.4K]

Remember
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2 years ago

Read 2 more answers

Find all of the equilibrium solutions. Enter your answer as a list of ordered pairs (R,W), where R is the number of rabbits and

zloy xaker [14]

Answer:

$(0,0)$ $(4000,0)$ and $(500,79)$

Step-by-step explanation:

Given

See attachment for complete question

Required

Determine the equilibrium solutions

We have:

$\frac{dR}{dt} = 0.09R(1 - 0.00025R) - 0.001RW$

$\frac{dW}{dt} = -0.02W + 0.00004RW$

To solve this, we first equate $\frac{dR}{dt}$ and $\frac{dW}{dt}$ to 0.

So, we have:

$0.09R(1 - 0.00025R) - 0.001RW = 0$

$-0.02W + 0.00004RW = 0$

Factor out R in $0.09R(1 - 0.00025R) - 0.001RW = 0$

$R(0.09(1 - 0.00025R) - 0.001W) = 0$

Split

$R = 0$ or $0.09(1 - 0.00025R) - 0.001W = 0$

$R = 0$ or $0.09 - 2.25 * 10^{-5}R - 0.001W = 0$

Factor out W in $-0.02W + 0.00004RW = 0$

$W(-0.02 + 0.00004R) = 0$

Split

$W = 0$ or $-0.02 + 0.00004R = 0$

Solve for R

$-0.02 + 0.00004R = 0$

$0.00004R = 0.02$

Make R the subject

$R = \frac{0.02}{0.00004}$

$R = 500$

When $R = 500$ , we have:

$0.09 - 2.25 * 10^{-5}R - 0.001W = 0$

$0.09 -2.25 * 10^{-5} * 500 - 0.001W = 0$

$0.09 -0.01125 - 0.001W = 0$

$0.07875 - 0.001W = 0$

Collect like terms

$- 0.001W = -0.07875$

Solve for W

$W = \frac{-0.07875}{ - 0.001}$

$W = 78.75$

$W \approx 79$

$(R,W) \to (500,79)$

When $W = 0$ , we have:

$0.09 - 2.25 * 10^{-5}R - 0.001W = 0$

$0.09 - 2.25 * 10^{-5}R - 0.001*0 = 0$

$0.09 - 2.25 * 10^{-5}R = 0$

Collect like terms

$- 2.25 * 10^{-5}R = -0.09$

Solve for R

$R = \frac{-0.09}{- 2.25 * 10^{-5}}$

$R = 4000$

So, we have:

$(R,W) \to (4000,0)$

When $R =0$ , we have:

$-0.02W + 0.00004RW = 0$

$-0.02W + 0.00004W*0 = 0$

$-0.02W + 0 = 0$

$-0.02W = 0$

$W=0$

So, we have:

$(R,W) \to (0,0)$

Hence, the points of equilibrium are:

$(0,0)$ $(4000,0)$ and $(500,79)$

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Harlamova29_29 [7]

Answer:

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