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2 years ago

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Find a solution to the following system of equations. –5x + y = –5 –4x + 2y = 2

1 answer:

Arte-miy333 [17]2 years ago

8 0

-5x + y = -5.....multiply by -2
-4x + 2y = 2
---------------
10x - 2y = 10 (result of multiplying by -2)
-4x + 2y = 2
-----------------add
6x = 12
x = 12/6
x = 2

-5x + y = -5
-5(2) + y = -5
-10 + y = -5
y = -5 + 10
y = 5

solution is (2,5)

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Answer:

The sample size is $n = 600$

Step-by-step explanation:

From the question we are told that

The sample proportion is $\r p = 0.48$

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Given that the confidence level is 95% the level of significance is mathematically represented as

$\alpha = 100 - 95$

$\alpha = 5 \%$

$\alpha = 0.05$

Next we obtain the critical value of $\frac{\alpha }{2}$ from the normal distribution table , the values is

$Z_{\frac{\alpha }{2} } = 1.96$

The reason we are obtaining critical value of $\frac{\alpha }{2}$ instead of $\alpha$ is because

$\alpha$ represents the area under the normal curve where the confidence level interval ( $1-\alpha$ ) did not cover which include both the left and right tail while

$\frac{\alpha }{2}$ is just the area of one tail which what we required to calculate the margin of error

Generally the margin of error is mathematically represented as

$MOE = Z_{\frac{\alpha }{2} } * \sqrt{ \frac{\r p(1- \r p )}{n} }$

substituting values

$0.04= 1.96* \sqrt{ \frac{0.48(1- 0.48 )}{n} }$

$0.02041 = \sqrt{ \frac{0.48(52 )}{n} }$

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3 years ago

Use the exponential decay model, Upper A equals Upper A 0 e Superscript kt, to solve the following. The half-life of a certai

Akimi4 [234]

Answer:

It will take 7 years ( approx )

Step-by-step explanation:

Given equation that shows the amount of the substance after t years,

$A=A_0 e^{kt}$

Where,

$A_0$ = Initial amount of the substance,

If the half life of the substance is 19 years,

Then if t = 19, amount of the substance = $\frac{A_0}{2}$ ,

i.e.

$\frac{A_0}{2}=A_0 e^{19k}$

$\frac{1}{2} = e^{19k}$

$0.5 = e^{19k}$

Taking ln both sides,

$\ln(0.5) = \ln(e^{19k})$

$\ln(0.5) = 19k$

$\implies k = \frac{\ln(0.5)}{19}\approx -0.03648$

Now, if the substance to decay to 78% of its original amount,

Then $A=78\% \text{ of }A_0 =\frac{78A_0}{100}=0.78 A_0$

$0.78 A_0=A_0 e^{-0.03648t}$

$0.78 = e^{-0.03648t}$

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$\ln(0.78) = -0.03648t$

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2 years ago

Andre45 [30]

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