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34kurt
3 years ago
12

Answer please im begging you QwQ

Mathematics
1 answer:
miskamm [114]3 years ago
4 0

Answer:

A is your answer

Step-by-step explanation:

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A veterinarian estimated the weight of a puppy to be 7 kg. The actual weight of the puppy was 6.6 kg.
KIM [24]

9514 1404 393

Answer:

  • absolute error: 0.4 kg
  • relative error: 6.1%

Step-by-step explanation:

The absolute error is the difference between the estimate and the actual weight:

  7.0 kg -6.6 kg = 0.4 kg . . . . absolute error

__

The relative error is the ratio of the absolute error to the actual weight:

  (0.4 kg)/(6.6 kg) × 100% = 6.060606...% ≈ 6.1% . . . . relative error

6 0
3 years ago
Help me plssssssssss
Dmitry_Shevchenko [17]

Answer:

just do the thing and then thing the do

Step-by-step explanation:

4 0
4 years ago
Please help ! thanks ! :)
Rufina [12.5K]

Answer:

a) is B because it inrease quite rapidly and then slowly increase

b) is c as it increase and then decreases quite soon and rapid and then increase quite rapidly again

Step-by-step explanation:

4 0
3 years ago
Which ordered pair is in the solution set y ≥1/3x+4
s2008m [1.1K]

the answe is ....

.....

-6,1

step by step

first you sympypfly the answer and then what you want to do is mutyplying te answer by the formula of the graph by then you should already have the table of the formula,then what you want to do is to try to coordinate the number of the graph,AND THA'S HOW YOU GET (-6,1)

4 0
3 years ago
Read 2 more answers
Find the limit. Use l'Hospital's Rule where appropriate. If there is a more elementary method, consider using it. lim x→9 x − 9
luda_lava [24]

Without resorting to L'Hopitâl's rule,

\displaystyle\lim_{x\to9}\frac{x-9}{x^2-81}=\lim_{x\to9}\frac{x-9}{(x-9)(x+9)}=\lim_{x\to9}\frac1{x+9}=\frac1{18}

With the rule, we get the same result:

\displaystyle\lim_{x\to9}\frac{x-9}{x^2-81}=\lim_{x\to9}\frac1{2x}=\frac1{18}

8 0
3 years ago
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