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2 years ago

15

a student did an experiment and determined the percent sugar in a piece of gum to be 67%. The accepted value for th percent suga

r in a piece of gum is 74%. what is the percent error in the students' calculations?

1 answer:

riadik2000 [5.3K]2 years ago

6 0

I don’t know, I need points

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For a normal distribution,

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Answer:

1.282, -0.5244,...

Step-by-step explanation:

We know that Z is a standard normal variate with mean =0, and std dev =1

a) z-score separates the highest 10% from the rest of the scores

= 90th percentile = 1.282

b) z-score separates the lowest 30% from the rest of the scores

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c) z-score separates the lowest 40% from the rest of the scores

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3 years ago

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585 is the answer of your question.

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3 years ago

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Abbie has to do 25 math problems tonight. She has completed 40%. How many problems has she completed?

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3 years ago

48 students are going to the zoo. one-fourth brought umbrellas. how many people did not bring umbrellas

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3 years ago

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PLEASE SHOW WORK

CaHeK987 [17]

(C)

Step-by-step explanation:

The volume of the conical pile is given by

$V = \dfrac{\pi}{3}r^2h$

Taking the derivative of V with respect to time, we get

$\dfrac{dV}{dt} = \dfrac{\pi}{3}\dfrac{d}{dt}(r^2h)$

$\:\:\:\:\:\:\:= \dfrac{\pi}{3}\left(2rh\dfrac{dr}{dt} + r^2\dfrac{dh}{dt}\right)$

Since r is always equal to h, we can set

$\dfrac{dr}{dt} = \dfrac{dh}{dt}$

so that our expression for dV/dt becomes

$\dfrac{dV}{dt} = \dfrac{\pi}{3}\left(3r^2\dfrac{dh}{dt}\right)$

$\:\:\:\:\:\:\:= \pi r^2\dfrac{dh}{dt}$

Solving for dh/dt, we get

$\dfrac{dh}{dt} = \dfrac{1}{\pi r^2}\dfrac{dV}{dt}$

$\:\:\:\:\:\:\:= \dfrac{1}{9\pi\:\text{m}^2}(36\:\text{m}^3\text{/s})$

$\:\:\:\:\:\:\:= \dfrac{4}{\pi}\:\text{m/s}$

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3 years ago