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inn [45]
3 years ago
5

Solvex2 - 8x + 12 =0 for x. The solutions for x are and

Mathematics
1 answer:
Marianna [84]3 years ago
3 0

Answer:

2 and 6

Step-by-step explanation:

First factor x from the expression

x^2-2x-6x+12=0

x(x-2)-6x + 12=0

Then factor out x-2 from the expression

x(x-2)-6x + 12=0

(x-2)(x-6)=0

When the product of factors equals 0, at least one factor is 0

(x-2)(x-6)=0

x-2=0

x-6-0

And so

X=2

X=6

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210s(9p/14s)=135p

So 135 seeds grew into plants.
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3 years ago
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After a shopping spree at Target, John realized he spent three times more money than his brother Mike. They spent a total of $16
PSYCHO15rus [73]

Answer:

John spent around $53

Step-by-step explanation:

The equation is 3x=160

Since the "x" is already by itself on one side, divide 160 by 3

160÷3=53.333⇒

Now if that's not the way IT CAN ALSO BE:

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3 years ago
A laboratory scale is known to have a standard deviation (sigma) or 0.001 g in repeated weighings. Scale readings in repeated we
weqwewe [10]

Answer:

99% confidence interval for the given specimen is [3.4125 , 3.4155].

Step-by-step explanation:

We are given that a laboratory scale is known to have a standard deviation (sigma) or 0.001 g in repeated weighing. Scale readings in repeated weighing are Normally distributed with mean equal to the true weight of the specimen.

Three weighing of a specimen on this scale give 3.412, 3.416, and 3.414 g.

Firstly, the pivotal quantity for 99% confidence interval for the true mean specimen is given by;

        P.Q. = \frac{\bar X - \mu}{\frac{\sigma}{\sqrt{n} } } ~ N(0,1)

where, \bar X = sample mean weighing of specimen = \frac{3.412+3.416+3.414}{3} = 3.414 g

            \sigma = population standard deviation = 0.001 g

            n = sample of specimen = 3

            \mu = population mean

<em>Here for constructing 99% confidence interval we have used z statistics because we know about population standard deviation (sigma).</em>

So, 99% confidence interval for the population​ mean, \mu is ;

P(-2.5758 < N(0,1) < 2.5758) = 0.99  {As the critical value of z at 0.5% level

                                                            of significance are -2.5758 & 2.5758}

P(-2.5758 < \frac{\bar X - \mu}{\frac{\sigma}{\sqrt{n} } } < 2.5758) = 0.99

P( -2.5758 \times {\frac{\sigma}{\sqrt{n} } } < {\bar X - \mu} < 2.5758 \times {\frac{\sigma}{\sqrt{n} } } ) = 0.99

P( \bar X-2.5758 \times {\frac{\sigma}{\sqrt{n} } } < \mu < \bar X+2.5758 \times {\frac{\sigma}{\sqrt{n} } } ) = 0.99

<u>99% confidence interval for</u> \mu = [ \bar X-2.5758 \times {\frac{\sigma}{\sqrt{n} } } , \bar X+2.5758 \times {\frac{\sigma}{\sqrt{n} } } ]

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                                             = [3.4125 , 3.4155]

Therefore, 99% confidence interval for this specimen is [3.4125 , 3.4155].

6 0
3 years ago
Write 300 + 90 + 4 + 0.5 + 0.07 in standard form
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Rewrite 6 x 6 x 6 x 6 using a exponent​
MAXImum [283]

Answer:

6^4

Step-by-step explanation:

There are four 6's.

6 0
3 years ago
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