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2 years ago

Solve for x 8 + x/3 = 15

Mathematics

2 answers:

Natali5045456 [20]2 years ago

8 0

Answer:

x = 21

Step-by-step explanation:

subtract 8 from both sides

you get x/3 = 7

multiply both sides by 3 you get x=21

Ulleksa [173]2 years ago

4 0

Answer:

Step-by-step explanation:

$8+\dfrac{x}{3}=15\qquad\text{multiply both sides by 3}\\\\(3)(8)+3\!\!\!\!\diagup^1\cdot\dfrac{x}{3\!\!\!\!\diagup_1}=(3)(15)\\\\24+x=45\qquad\text{subtract 24 from both sides}\\\\24-24+x=45-24\\\\x=21$

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The greater of two consecutive integers is 20 more than twice the smaller. find the integers.

Kipish [7]

Let x represent the smaller. Then x+1 is the greater of the two.

... x+1 = 2x +20

... 0 = x + 19 . . . . . subtract x+1

... x = -19

Your two integers are -19 and -18.

7 0

3 years ago

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nasty-shy [4]

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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> $\mu$ = [ $\bar X-2.5758 \times {\frac{\sigma}{\sqrt{n} } }$ , $\bar X+2.5758 \times {\frac{\sigma}{\sqrt{n} } }$ ]

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