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pashok25 [27]
3 years ago
3

What is the solution to the equation StartFraction 1 Over h minus 5 EndFraction + StartFraction 2 Over h + 5 EndFraction = Start

Fraction 16 Over h squared minus 25 EndFraction? h = eleven-thirds h = 5 h = 7 h = twenty-one-halves
Mathematics
2 answers:
jonny [76]3 years ago
5 0

Answer:

h=7

Step-by-step explanation:

we have

\frac{1}{h-5}+\frac{2}{h+5}=\frac{16}{h^2-25}

Solve for h

Remember that

h^2-25=(h-5)(h+5) ----> by difference of squares

Multiply by  h^2-25  both sides to remove the fractions

(h+5)+2(h-5)=16\\3h-5=16\\3h=21\\h=7

Natali5045456 [20]3 years ago
3 0

Answer:

C. h = 7

Step-by-step explanation:

What is the solution to the equation StartFraction 1 Over h minus 5 EndFraction + StartFraction 2 Over h + 5 EndFraction = StartFraction 16 Over h squared minus 25 EndFraction?

h = eleven-thirds

h = 5

h = 7

h = twenty-one-halves

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3 years ago
The measure of angle is 5. The equivalent measurement in degrees is... ​
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300 degrees

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2 years ago
On Sunday, Sheldon bought 412 kg of plant food. He used 123kg on his strawberry plants and used 14 kg for his tomato plants.
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Answer:

275

Step-by-step explanation:

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3 years ago
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Because, Y, which is the shaded part on the graph

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6 0
4 years ago
The useful life of a radial tire is normally distributed with a mean of 30,000 miles and a standard deviation of 5000 miles. The
otez555 [7]

Answer:

  empirical rule: 81.5%

  table or calculator: 81.9%

Step-by-step explanation:

The lower limit of the life range of interest has a z-score of ...

  z = (x -μ)/σ = (20,000 -30,000)/5,000 = -2

The upper limit has a z-score of ...

  z = (35,000 -30,000)/5,000 = 1

<u>Empirical rule solution</u>

The empirical rule tells you that 95% of the distribution lies within 2 standard deviations of the mean, so (100% -95%)/2 = 2.5% lie below z = -2. It also tells you 68% lie within 1 standard deviation of the mean, so (100% -68%)/2 = 16% lie above z = 1.

The fraction that lies within -2 to 1 standard deviations of the means is thus ...

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The probability the tire has a life in the desired range is about 81.5%.

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<u>Calculator solution</u>

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  P(-2 < z < 1) ≈ 0.8185946

The probability the tire has a life in the desired range is about 81.9%.

6 0
3 years ago
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