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3 years ago

The area of a blackboard is

Mathematics

2 answers:

noname [10]3 years ago

8 0

Answer:

3/2

Step-by-step explanation:

unit rate describes how many units of the first type of quantity corresponds to one unit of the second type of quantity so:

1.1/3= 3.1+1÷3=4/3

8/9÷4/3=3/2

Sonja [21]3 years ago

7 0

Answer:

<h2>There are 6 square yards of blackboard per 2 square yards per poster.</h2>

Step-by-step explanation:

The unit rate refers to how many area units of the poster are in the blackboard.

To find this rate, we just have to divide the area of the blackboard by the area of the poster.

$B=1\frac{1}{3}=\frac{4}{3}$

Where $B$ represents the area of the blackboard in square yards.

$P=\frac{8}{9}$

Where $P$ represents the area of the poster.

Now, the ratio is

$r= \frac{B}{P}=\frac{4}{3} \div \frac{8}{9}=\frac{4}{3} \times \frac{9}{8}=\frac{36}{24}\\r=\frac{6}{4}=\frac{3}{2}=1.5$

This mean that the unit rate is 1.5, that is, there are 6 square yards of blackboard per 2 square yards per poster.

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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

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

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