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3 years ago

13

On a math test, Janice answers 85% of the questions correctly. If she answered 17 questions correctly, how many questions are on

the test?
Please help me

1 answer:

alex41 [277]3 years ago

5 0

Answer:

20

Step-by-step explanation:

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Kenya exchanges $200 for euros (€). Suppose the conversion rate is €1 = $1.321. How many euros should Kenya receive? (1 point) Q

Paul [167]

Answer:

151.4€

Step-by-step explanation:

1€ = $1.321

$200 = 200/1.321 = 151.4€

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3 years ago

Values that are equivalent to 0.8

-Dominant- [34]

0,8 = 8/10 = 4/5 = 16/20...

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3 years ago

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4/9 divided by 1 and 7/15

FromTheMoon [43]

I believe it’s 10/33 because you divide 4/9 and 22/15 and use kfc you get 60/198 simplifying it makes 10/33

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3 years ago

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What has the same value as |-5| ?

Georgia [21]

Answer:

5

Step-by-step explanation:

This is because |-5| is equal to 5

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3 years ago

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A wooden artifact from an ancient tomb contains 20 percent of the carbon-14 that is present in living trees. How long ago, to th

maxonik [38]

Answer:

The artifact was made approximately 13,306 years ago

Step-by-step explanation:

The half-life of Carbon-14 = 5730 years

The formula for radioactive decay is given by the equation:

$N=N_0 e^{- \lambda t}\\where:\\N = Number\ of\ particles\ at\ time\ t\\N_0= Number\ of\ particles\ at\ time\ t_0\\\lambda = decay\ constant = \frac{In}{t_{1/2}} ; t_{1/2} = half-life\\t = time\ in\ years$

The amount of Carbon-14 left = 20% = 20/100 = 0.2

$\therefore \frac{N}{N_0} = 20 \% = 0.2$

$N=N_0 e^{- \lambda t}\\\\dividing\ both\ sides\ by\ N_0\\\frac{N}{N_0} = e^{- \lambda t}\\0.2 = e^{- \lambda t}\\taking\ natural\ logarithm\ of\ both\ sides\\In 0.2 = Ine^{- \lambda t}\\In0.2 = -\lambda \times t\\but\ \lambda = \frac{In2}{t_{1/2}} \\\therefore In0.2 = -\lambda \times t\\\\= In0.2 = - \frac{In2}{t_{1/2}} \times t\\where\ t_{1/2} = 5730\ years\\-1.6094 = - \frac{0.6931}{5730} \times t\\-1.6094 = - 0.00012095 \times t\\t = \frac{-1.6094}{-0.00012095} \\t = 13,306.32$

t ≈ 13,306 years

7 0

2 years ago