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tamaranim1 [39]

3 years ago

14

recycling one metric ton of corrugated cardboard will sAVE ABOUT 17 TREES. a WAREHOUSE RECYCLED 14,327 KILOGRAMS OF CORRUGATED C

ARDBOARD OVER THE LAST YEAR. hOW MANY TREES DID THIS SAVE.

2 answers:

melomori [17]3 years ago

8 0

Answer:

243 trees were saved.

Step-by-step explanation:

Recycling one metric ton (1000 kg) of corrugated cardboard will save number of trees = 17

Warehouse recycled 14327 kg of cardboard last year. So we have to find the number of trees saved.

To find the number of trees we will use the unitary method.

∵ Recycling 1000 kg of cardboard number of trees saved = 17

∴ Recycling 1 kg cardboard number of trees saved = $\frac{17}{1000}$

∴ 14327 kg cardboard will save the number of trees = $\frac{17}{1000}\times 14327$

= 243.559

≈ 243 trees

Therefore, 243 trees were saved.

slega [8]3 years ago

3 0

1 kg = 0.001 metric tons..so 14327 kg = (14327 * 0.001) = 14.327 metric tons
and if 1 metric tons saves 17 trees....then 14.327 tons saves : 14.327 * 17 = 243.559 saved trees...and since u cant have 0.559 trees, u would round to 243 trees

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Step-by-step explanation:

From a population (Bernoulli population), 90% of the bearings meet a thickness specification, let $p_1$ be the probability that a bearing meets the specification.

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Variance: $\sigma_1^2=n_1p_1(1-p_1)=500\times0.9\times0.1=45$

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So, $X\geq 440.$

Here, 440 is included, so, by using the continuity correction, take x=439.5 to compute z score for the normal distribution.

$z=\frac{x-\mu}{\sigma}=\frac{339.5-450}{6.71}=-1.56$ .

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$P(z\geq-1.56)=\int_{-1.56}^{\infty}\frac{1}{\sqrt{2\pi}}e^{\frac{-z^2}{2}}=0.94062$

Hence, the probability that a given shipment is acceptable is 0.94.

(b) We have the probability of acceptability of one shipment 0.94, which is same for each shipment, so here the number of shipments is a Binomial population.

Denote the probability od acceptance of a shipment by $p_2$ .

$p_2=0.94$

The total number of shipment, i.e sample size, $n_2= 300$

Here, the sample size is sufficiently large to approximate it as a normal distribution, for which mean, $\mu_2$ , and variance, $\sigma_2^2$ .

Mean: $\mu_2=n_2p_2=300\times0.94=282$

Variance: $\sigma_2^2=n_2p_2(1-p_2)=300\times0.94(1-0.94)=16.92$

$\Rightarrow \sigma_2=\sqrt(16.92}=4.11$ .

In this case, X>285, so, by using the continuity correction, take x=285.5 to compute z score for the normal distribution.

$z=\frac{x-\mu}{\sigma}=\frac{285.5-282}{4.11}=0.85$ .

So, the probability that a given shipment is acceptable is

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Hence, the probability that a given shipment is acceptable is 0.20.

(c) For the acceptance of 99% shipment of in the total shipment of 300 (sample size).

The area right to the z-score=0.99

and the area left to the z-score is 1-0.99=0.001.

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$-3.09=\frac{285.5-300\alpha}{\sqrt{300 \alpha(1-\alpha)}}$

On solving

$\alpha= 0.977896$

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Let $p$ be the probability that one bearing meets the specification. So

$-2.01=\frac{439.5-500 p}{\sqrt{500 p(1-p)}}$

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