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3 years ago

Which of the following scatter plots below show the most accurate line of best fit

Mathematics

1 answer:

dezoksy [38]3 years ago

3 0

Answer:

The one in the top right

Step-by-step explanation:

All of the points seem to match the line best, each of the other graphs vary more.

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Please help on this one?

Sedaia [141]

The inverse is where the x and y values are flipped, so the left side would have 6 7 8 9 and the right side would have 9 10 11 12 for the inverse.

Therefore your answer is A.

4 0

3 years ago

Square ABCD is shown with four congruent images such that ABCD FGHI JKLM NOPQ RSTU.

N76 [4]

FGHI

Step-by-step explanation:

The square FGHI needs to be created into a reflection for true congruence of the figure thus transformed across the axis.

FGHI's co-ordinates' signs are to be reversed for reflection thus coordinates:

2,-5

4,-5

2,-7

4,-7

are made into

-2,5

-4,5

-2,7

-4,7

5 0

3 years ago

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I'm lazy but have a lot of points

ohaa [14]

Answer: 37

Step-by-step explanation:

Well DCB is a right angle, and there is already half which is 53 degrees, subtract 90 by 53 and you will get your answer.

4 0

3 years ago

Initially 100 milligrams of a radioactive substance was present. After 6 hours the mass had decreased by 3%. If the rate of deca

Hitman42 [59]

Answer:

The half-life of the radioactive substance is 135.9 hours.

Step-by-step explanation:

The rate of decay is proportional to the amount of the substance present at time t

This means that the amount of the substance can be modeled by the following differential equation:

$\frac{dQ}{dt} = -rt$

Which has the following solution:

$Q(t) = Q(0)e^{-rt}$

In which Q(t) is the amount after t hours, Q(0) is the initial amount and r is the decay rate.

After 6 hours the mass had decreased by 3%.

This means that $Q(6) = (1-0.03)Q(0) = 0.97Q(0)$ . We use this to find r.

$Q(t) = Q(0)e^{-rt}$

$0.97Q(0) = Q(0)e^{-6r}$

$e^{-6r} = 0.97$

$\ln{e^{-6r}} = \ln{0.97}$

$-6r = \ln{0.97}$

$r = -\frac{\ln{0.97}}{6}$

$r = 0.0051$

$Q(t) = Q(0)e^{-0.0051t}$

Determine the half-life of the radioactive substance.

This is t for which Q(t) = 0.5Q(0). So

$Q(t) = Q(0)e^{-0.0051t}$

$0.5Q(0) = Q(0)e^{-0.0051t}$

$e^{-0.0051t} = 0.5$

$\ln{e^{-0.0051t}} = \ln{0.5}$

$-0.0051t = \ln{0.5}$

$t = -\frac{\ln{0.5}}{0.0051}$

$t = 135.9$

The half-life of the radioactive substance is 135.9 hours.

6 0

3 years ago

What is the base higher and area

algol13

Answer:

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers