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3 years ago

For what values of h and k does the linear system have infinitely many solutions?

Mathematics

1 answer:

Sophie [7]3 years ago

7 0

Answer:

h=-32 and k=-24

Step-by-step explanation:

$4x_1 + 3x_2 = -1:a_1x_1 + b_1x_2=c_1\\hx_1 +kx_2=8:a_2x_1+b_2x_2=c_2$

For infinitely many solutions, lines must be same; $\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}\\\\=\frac{4}{h}=\frac{3}{k}=\frac{-1}{8}\\\\h=-32, k=-24$

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2 years ago

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MrRissso [65]

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A) sample mean = $1.36 million

B) standard deviation = $0.9189 million

C) confidence interval = ($1.93 million , $0.79 million)

*since the sample size is very small, the confidence interval is not valid.

Step-by-step explanation:

samples:

$2.7 million
$2.4 million
$2.2 million
$2 million
$1.5 million
$1.5 million
$0.5 million
$0.5 million
$0.2 million
$0.1 million

sample mean = $1.36 million

the standard deviation:

$2.7 million - $1.36 million = 1.34² = 1.7956
$2.4 million - $1.36 million = 1.04² = 1.0816
$2.2 million - $1.36 million = 0.84² = 0.7056
$2 million - $1.36 million = 0.64² = 0.4096
$1.5 million - $1.36 million = 0.14² = 0.0196
$1.5 million - $1.36 million = 0.14² = 0.0196
$0.5 million - $1.36 million = -0.86² = 0.7396
$0.5 million - $1.36 million = -0.86² = 0.7396
$0.2 million - $1.36 million = -1.16² = 1.3456
$0.1 million - $1.36 million = -1.26² = 1.5876
total $8.444 million / 10 = $0.8444 million

standard deviation = √0.8444 = 0.9189

95% confidence interval = mean +/- 1.96 standard deviations/√n:

$1.36 million + [(1.96 x $0.9189 million)/√10] = $1.36 million + $0.57 million = $1.93 million

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