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HACTEHA [7]
3 years ago
6

In 2 yrs time ella will be twice the age she was 5 yrs ago. how old is she now?

Mathematics
1 answer:
chubhunter [2.5K]3 years ago
7 0
8 because 2 times 5 is ten
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PLEASE I NEED HELP ASAP! The function below represents the interest Jessi earns on an investment. Identify the term that represe
luda_lava [24]

Answer:

1000

Step-by-step explanation!

The formula for the amount accrued [ƒ(x)] on an investment earning compound interest is f(t) = P(1 + r)^t where:

P = the amount of money invested (the principal)

r = the interest rate per payment period expressed as a decimal fraction

t = the number of periods

Your formula is

f(x) = 1000(1 + 0.05)^x

In comparison, we can see that the term that represents the amount of money originally invested is 1000.

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Which of these expressions is equivalent to 3/6
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1/2, 6/12, 9,/18, and 15/30
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Elena has $5 for pencils. Each pencil costs $0.30. How many pencils can she buy? *No run-on decimals please!*
aleksley [76]

Answer:

she could by 16 pencils with 20 cents left over

Step-by-step explanation:

so the fraction is 16 and 2/3

so she could buy 16 pencils and have 20 cents leftover

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2 years ago
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Jeremy is a 17-year-old junior in high school and just started his first job. He wants to open a savings account. Which of these
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. A shipyard makes a container ship that can withstand the total amount of weight W, which is normally distributed with mean of
Debora [2.8K]

Answer:

the maximum number of containers that the ship can load is 170

Step-by-step explanation:

Given the data in the question;

W ~ N( 600, 60 )

S ~ N( 4n, 0.4√n )

so

p( W > S ) = 0.90

⇒ P( W - S > 0) = 0.9 ------ let this be equation 1

now, since W and S are independent

Mean( W - S ) = Mean( W ) - Mean( S 0 = 600 - 4n

and

SD( W - S ) = √( var(W) + var(S) ) = √( 60² + 0.4²n)

hence;

W - S ~ N( 600 - 4n, √( 60² + 0.4²n) )

now, from equation one, P( W - S > 0) = 0.9

P( \frac{(W-S)-Mean(W-S)}{SD(W-S)} > \frac{0-(600-4n)}{\sqrt{60^2 + 0.4^2n} }) = 0.90

P( z > \frac{0-(600-4n)}{\sqrt{60^2 + 0.4^2n} }) = 0.90

from z- table

P( z > \frac{0-(600-4n)}{\sqrt{60^2 + 0.4^2n} }) = P( z >-1.282)  

\frac{4n - 600}{\sqrt{60^2 + 0.4^2n} } = -1.282 ------------------ let this be equation 2

now, we square both sides of equation 2

\frac{(4n - 600)^2}{60^2 + 0.4^2n} } = (-1.282)^2  

\frac{(4n - 600)(4n-600)}{3600 + 0.16n} } = 1.643524

we cross multiply

16n² + 360000 - 4800n = 1.643524( 3600 + 0.16n )

16n² + 360000 - 4800n = 5916.6864 + 0.26296384n

16n² + 360000 - 5916.6864 - 4800n - 0.26296384n = 0

16n² + 354083.3136 - 4800.26296384n = 0    

16n² - 4800.26296384n + 354083.3136 = 0  

solving the quadratic equation, we know that;

x = -b±√( b² - 4ac ) / 2a

so we substitute

x = [-(-4800.26296384) ±√( (-4800.26296384)² - (4 × 16 × 354083.3136)] / [2×16]

x = [ 4800.26296384 ±√( 23042524.522 - 22661332.0704 ] / 32

x = [ 4800.26296384 ±√(381192.4516) ] / 32

x = [ 4800.26296384 ± 617.4078 ] / 32

Hence;

x = [ 4800.26296384 - 617.4078 ] / 32 or  [ 4800.26296384 + 617.4078 ] / 32

x = 131   or  170

Therefore, the maximum number of containers that the ship can load is 170

5 0
3 years ago
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