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3 years ago

Is n+4 equal to 8(n - 4)

Mathematics

2 answers:

Dafna11 [192]3 years ago

5 0

Yes. its correct. you set them equal to get the answer. then plus that answer into both. i have sloppy writing but look

weqwewe [10]3 years ago

4 0

No, n+4=4n an 8(n-4)=32n

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The age of a father is 2 less than 7 times the age of his son. In 3 years, the sum of their ages will be 52. If the son’s presen

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Answer:

First,

the age of the son: s

the age of the father is 2 less than 7 times the age of his son: 7s - 2

In 3 years, the sum of their ages will be 52, In 3 years, the age of the sun will

increase 3, so s+3. Also, the age of his

father also increase 3, (7s-2)+3

(s+3)+ [(7s-2)+3] = 52 is the equation.

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2 years ago

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3 years ago

HELP If 90 containers of candy hearts cost $2.70, what is the constant of proportionality

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3 years ago

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4 years ago

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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

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

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

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

= [ $3.414-2.5758 \times {\frac{0.001}{\sqrt{3} } }$ , $3.414+2.5758 \times {\frac{0.001}{\sqrt{3} } }$ ]

= [3.4125 , 3.4155]

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

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3 years ago