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1 year ago

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FS Math FS LV2 Volume Problems Q1. The swimming pool is in the shape of a cuboid 20 metres long and 8 metres wide. It is filled

with water to a depth of 1.8 metres. Diagram not accurately drawn 1.8m 8 m 20m Charles needs to add chlorine powder to the water in the pool to keep the water clean. Each week Charles needs to use 300 grams of chlorine powder per 50 m of water. Chlorine powder is sold in boxes. Each box contains 5000 grams of chlorine powder. Charles thinks that one box will be enough for 4 weeks. Is one box enough for 4 weeks? Show why you think this.
pls answer

1 answer:

DedPeter [7]1 year ago

3 0

From the volume computed, since each box contains 5000 grams of chlorine powder, it won't be enough for 4 weeks.

<h3>How to calculate the volume</h3>

The volume of the swimming pool water will be:

= length × width × height.

= 20 × 8 × 1.8

= 288m³

Since 50m³ of water chlorine needs 300gm, therefore, 1m³ will need;

= 300/50 = 6

Therefore, 288m³ will need:

= 288 × 6 = 1728gm

The amount of chlorine powder needed for 4 weeks will be:

= 1728 × 4

= 6912 gm

Since each box contains 5000 grams of chlorine powder, it won't be enough for 4 weeks.

Learn more about volume on:

brainly.com/question/1972490

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Answer:

z=1.59

If we compare the p value and the significance level given $\alpha=0.025$ we see that $p_v>\alpha$ so we can conclude that we fail to reject the null hypothesis, so we can conclude that the mean repellency of the new bug repellent is greater than 89% at 0.025 of signficance.

Step-by-step explanation:

1) Data given and notation

$\bar X=0.92$ represent the mean effectiveness of a new bug repellent for the sample

$\sigma=0.08$ represent the population standard deviation for the sample

$n=18$ sample size

$\mu_o =0.89$ represent the value that we want to test

$\alpha=0.025$ represent the significance level for the hypothesis test.

t would represent the statistic (variable of interest)

$p_v$ represent the p value for the test (variable of interest)

State the null and alternative hypotheses.

We need to conduct a hypothesis in order to determine if the mean repellency of the new bug repellent is greater than 89% or 0.89, the system of hypothesis would be:

Null hypothesis: $\mu \leq 0.89$

Alternative hypothesis: $\mu > 0.89$

We don't know the population deviation, and the sample size <30, so for this case is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$ (1)

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".

Calculate the statistic

We can replace in formula (1) the info given like this:

$t=\frac{0.92-0.89}{\frac{0.08}{\sqrt{18}}}=1.59$

Calculate the P-value

First we need to calculate the degrees of freedom given by:

$df=n-1=18-1=17$

Since is a one-side upper test the p value would be:

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In Excel we can use the following formula to find the p value "=1-T.DIST(1.59;17;TRUE)"

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Answer:

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Now Substituting equation 1 in equation 2 we get;

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