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2 years ago

Express each fraction or mixed number as a percent

1 answer:

vladimir1956 [14]2 years ago

3 0

So,

#1: $\frac{5}{8}$

We know that $\frac{1}{8}$ = 0.125, or 12.5%.
Therefore, we just have to multiply 12.5% by 5 to get $\frac{5}{8}$ in percent form.

12.5(5) = 62.5

$\frac{5}{8}$ as a percent is 62.5%.

#2: $2 \frac{4}{7}$

$2 \frac{4}{7} = 200%$ + $\frac{4}{7}$

4/7 rounded to the nearest thousandth is 0.5714, or 57.14%.

200% + 57.14% = 257.14%

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HELP!!!? PLS ASP!!!!!!!

vovikov84 [41]

Answer:

Step-by-step explanation:

12x12 = 144 per side

144x6 sides = 864

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2 years ago

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Find the value of x.

UNO [17]

Answer:

x = 20

Step-by-step explanation:

Intersecting Chords Theorem: ab = cd

Step 1: Label our variables

a = x

b = x - 11

c = x - 8

d = x - 5

Step 2: Plug into theorem

x(x - 11) = (x - 5)(x - 8)

Step 3: Solve for x

x² - 11x = x² - 8x - 5x + 40

x² - 11x = x² - 13x + 40

-11x = -13x + 40

2x = 40

x = 20

5 0

3 years ago

Read 2 more answers

Can you please help me?

Igoryamba

The correct answer should be 192.
3(8)^2=3x64

6 0

2 years ago

A forest covers 31,000 acres. A survey finds that 0.2 % of the forest is old-growth trees. How many acres of old-growth trees

Oksana_A [137]

Answer:

There are 62 acres of old growth trees

Step-by-step explanation:

You would need to find 0.2% of 31,000

Lets say 0.2% of 31,000 is Y

Y = 0.2/100

Y = 0.2 ÷ 100 x 31,000

Y= 62

8 0

3 years ago

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