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3 years ago

7

Find the sum. 5 2a a? + 2a + 1 + a2 + 4a + 3

2 answers:

Vinvika [58]3 years ago

7 0

Answer:

$= \frac{2 {a}^{2} + 11a + 5}{ {(a + 1)}^{2}(a + 3)} \\$

Step-by-step explanation:

$\frac{2a}{ {a}^{2} + 2a + 1} + \frac{5}{ {a}^{2} + 4a + 3 } \\ \frac{2a}{(a + 1)(a + 1)} + \frac{5}{(a + 1)(a + 3)} \\ \frac{2a}{ {(a + 1)}^{2} } + \frac{5}{(a + 1)(a + 3)} \\ \frac{2a(a + 3) + 5(a + 1)}{ {(a + 1)}^{2}(a + 3) } \\ \frac{2 {a}^{2} + 6a + 5a + 5 }{ {(a + 1)}^{2}(a + 3)} \\ = \frac{2 {a}^{2} + 11a + 5}{ {(a + 1)}^{2}(a + 3)}$

Salsk061 [2.6K]3 years ago

5 0

$\frac{2a}{a^2+2a+1}+\frac{5}{a^2+4a+3}$

Factor the denominators.

$\frac{2a}{\left(a+1\right)^2}+\frac{5}{\left(a+1\right)\left(a+3\right)}$

Adjust fractions based on LCM.

$\frac{2a\left(a+3\right)}{\left(a+1\right)^2\left(a+3\right)}+\frac{5\left(a+1\right)}{\left(a+1\right)^2\left(a+3\right)}$

Denominators are same, so add the fractions.

$\frac{2a\left(a+3\right)+5\left(a+1\right)}{\left(a+1\right)^2\left(a+3\right)}$

Expand the numerator.

$\frac{2a^2+11a+5}{\left(a+1\right)^2\left(a+3\right)}$

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shusha [124]

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Step-by-step explanation:

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3 years ago

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3 years ago

The weights of items produced by a company are normally distributed with a mean of 4.5 ounces and a standard deviation of 0.3 ou

GalinKa [24]

Answer:

The right solution is:

(a) 0.8849

(b) 12.28%

(c) 4.9935

(d) 28004

Step-by-step explanation:

Given:

Mean,

= 4.5

Standard deviation,

= 0.3

(a)

P(x > 4.14)

As we know,

⇒ $z = \frac{4.14-4.5}{0.3}$

$=-1.20$

then,

⇒ $P(z>-1.20) = P(z$

$=0.8849$

(b)

P(4.8 < x < 5.04)

= $P(\frac{4.8-4.5}{0.3} < \frac{x-\mu}{\sigma} < \frac{5.04-4.5}{0.3} )$

= $P(1$

= $P(z$

= $0.9641 -0.8413$

= $0.1228$

or,

= $12.28$ (%)

(c)

P(x > x) = 0.05

z value will be,

= 1.645

⇒ $1.645 = \frac{x - 4.5}{0.3}$

$x = 4.9935$

(d)

P(x < 5.01)

⇒ $z = \frac{x- \mu}{\sigma}$

$=\frac{5.01-4.5}{0.3}$

$=1.7$

P(z < 1.70) = 0.9554

⇒ $n = \frac{27875}{0.9954}$

$=28004$

3 0

3 years ago