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hodyreva [135]
3 years ago
11

What is 4xy - 5y² - 3x² from 5x + 3y² - xy ? ​

Mathematics
2 answers:
andre [41]3 years ago
6 0

Refer the attachment for your answer

geniusboy [140]3 years ago
4 0

▪▪▪▪▪▪▪▪▪▪▪▪▪  {\huge\mathfrak{Answer}}▪▪▪▪▪▪▪▪▪▪▪▪▪▪

The equivalent expression is ~

\boxed{ \sf8 {y}^{2}  + 3 {x}^{2}  + 5x - 5xy}

\large \boxed{ \mathfrak{Step\:\: By\:\:Step\:\:Explanation}}

Let's solve ~

  • 5x² + 3 {y}^{2}  - xy - (4xy - 5y {}^{2}  - 3 {x}^{2} )

  • 5x² + 3 {y}^{2}  - xy - 4xy  +5 {y}^{2}  + 3 {x}^{2}

  • 3 {y}^{2}  + 5 {y}^{2}  + 3 {x}^{2} + 5x²  - xy - 4xy

  • 8 {y}^{2}  + 8 {x}^{2}   - 5xy
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Answer:

0.6563 or 65.63% of brook trout caught will be between 12 and 18 inches

Step-by-step explanation:

Mean trout length (μ) = 14 inches

Standard deviation (σ) = 3 inches

The z-score for any given trout length 'X' is defined as:  

z=\frac{X-\mu}{\sigma}  e interval

For a length of X =12 inches:

z=\frac{12-14}{3}\\z=-0.6667

According to a z-score table, a score of -0.6667 is equivalent to the 25.25th percentile of the distribution.

For a length of X =18 inches:

z=\frac{18-14}{3}\\z=1.333

According to a z-score table, a score of 1.333 is equivalent to the 90.88th percentile of the distribution.

The proportion of trout caught between 12 and 18 inches, assuming a normal distribution, is the interval between the equivalent percentile of each length:

P(12\leq X\leq 18) = 90.88\% - 25.25\%\\P(12\leq X\leq 18) = 65.63\%

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Answer:

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Step-by-step explanation:

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Step-by-step explanation:

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