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oksano4ka [1.4K]
4 years ago
9

A researcher finds that white bread contains more preservatives and has a higher moisture content than wheat bread. She designs

a test to determine whether the type of bread, white or wheat, affects the amount of mold growth. What must be done for the researcher to accurately measure the dependent variable?
A. She will need to allow mold to grow for more days on the white bread, since it contains more preservatives, to measure the dependent variable.
B. She will need to limit the dependent variable to white bread or wheat bread.
C. She will need to allow mold to grow for fewer days on the white bread, since it contains more preservatives, to measure the dependent variable.
D. She will need to learn out how to quantitatively measure mold growth to measure the dependent variable.

(I put 99 points on this; and you'll receive brainliest).
Physics
2 answers:
nekit [7.7K]4 years ago
7 0

D.

is correct on edg.

Crank4 years ago
6 0
For the researcher to accurately measure the dependent variable [which is quantity of mold growth produced] she will need to learn how to quantitatively measure mold growth. The correct option is D. Since the researcher has decided to measure the quantity of mold growth produced by both white and wheat bread, she must devise an effective and accurate mean of measuring the quantity of growth produced by the breads, this will enable her to subject the two breads to the same experimental conditions.
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You have 2 resistors of unknown values you label Ra and Rb. You have an old battery and a multimeter you bought years ago for 7$
Veseljchak [2.6K]

Answer:

the answers, the correct one is D,   Rb₂ = 29.97 ohm

Explanation:

For this exercise we use ohm's law and the equivalent resistance ratio for series and parallel circuits.

Serial circuit

         (Ra + Rb) is = V

         (Ra + Rb) 0.111 = 10

         (Ra + Rb) = 10 / 0.111 = 90.09

parallel circuit

         \frac{1}{R} = \frac{1}{Ra} + \frac{1}{Rb}

           R = \frac{Ra \ Rb}{Ra + Rb}

          \frac{Ra \ Rb}{Ra + Rb} i_p = V

         \frac{Ra \ Rb}{Ra + Rb} 0.5 = 10

         \frac{Ra \ Rb}{Ra + Rb} = 10 / 0.5 = 20

we write and solve our system of equations

         Ra + Rb = 90.09

         \frac{Ra \ Rb}{Ra + Rb} = 20

         

we solve for Ra in the first equation

          Ra = 90.09 - Rb

           RaRb = 20 (Ra + Rb)

we substitute Ra in the second equation

           (90.09-Rb) Rb = 20 [(90.09-Rb) + Rb]

            90.09 Rb - Rb² = 20 90.09

           

            Rb² - 90.09 Rb + 1801.8 = 0

we solve the quadratic equation

            Rb = [90.09 ±\sqrt{90.09^2 - 4 \ 1801.8} ] / 2

            Rb = [90.09 ± 30.15] / 2

            Rb₁ = 60.12 ohm

            Rb₂ = 29.97 ohm

the smallest value is Rb = 30 ohm

When checking the answers, the correct one is D

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Answer:

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