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oee [108]
3 years ago
7

A swimming pool is filled with water by using two taps A and B. Alone, it takes tap A 3 hours less than B to fill the same pool.

Together, they take 2 hours to fill the pool. How many hours does it take each tap to fill the swimming pools separately?
Mathematics
1 answer:
iris [78.8K]3 years ago
5 0

Answer:

Tap A 3hrs

Tap B 6hrs

Step-by-step explanation:

Let the volume of the swimming pool be Xm^3.

Now, to get the appropriate volume, we know we need to multiply the rate by the time. Let the rate of the taps be R1 and R2 respectively, while the time taken to fill the swimming pool be Ta and Tb respectively.

x/Ta= Ra

x/Tb= Rb

X/(Ra + Rb)= 2

Ta = Tb - 3

From equation 2:

X = 2( Ra + Rb)

Substituting the values of Ra and Rb Using the first set of equations

X = 2( x/Ta + x/Tb)

But Ta = Tb - 3

1/2 = 1/(Tb - 3)+ 1/Tb

0.5 = (Tb + Tb-3)/Tb(Tb - 3)

At this juncture let’s say Tb = y

0.5 = (2y - 3)/y(y - 3)

y(y-3 ) = 4y - 6

y^2 -3y - 4y + 6 = 0

y^2 -7y + 6= 0

Solving the quadratic equation, we get y =

y = Tb = 6hrs or 1hr

We remove one hour as we know that Tap A takes 3hrs left than tap B and there is nothing like negative hours

Now, we get Ta by Tb -3 = 6 - 3 = 3hrs

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A survey of 700 adults from a certain region​ asked, "What do you buy from your mobile​ device?" The results indicated that 59​%
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Answer:

a) the test statistic z = 1.891

the null hypothesis accepted at 95% level of significance

b) the critical values of 95% level of significance is zα =1.96

c) 95% of confidence intervals are  (0.523 ,0.596)

Step-by-step explanation:

A survey of 700 adults from a certain region

Given sample sizes n_{1} = 400 and n_{2} = 300

Proportion of mean p_{1} = \frac{236}{400} = 0.59 and p_{2} = \frac{156}{300} = 0.52

<u>Null hypothesis H0</u> : assume that there is no significant difference between males and women reported they buy clothing from their mobile device

p1 = p2

<u>Alternative hypothesis H1:</u>- p1 ≠ p2

a) The test statistic is

Z = \frac{p_{1} -p_{2} }{\sqrt{pq(\frac{1}{n_{1} }+\frac{1}{n_{2} )}  } }

where p = \frac{n_{1}p_{1} +n_{2}p_{2} }{n_{1}+n_{2}}= \frac{400X0.59+300X0.52}{700}  

on calculation we get   p = 0.56    

now q =1-p = 1-0.56=0.44

Z = \frac{p_{1} -p_{2} }{\sqrt{pq(\frac{1}{n_{1} }+\frac{1}{n_{2} )}  } }\\   =\frac{0.56-0.52}{\sqrt{0.56X0.44}(\frac{1}{400}+\frac{1}{300}   }

after calculation we get z = 1.891

b) The critical value at 95% confidence interval zα = 1.96 (from z-table)

The calculated z- value < the tabulated value

therefore the null hypothesis accepted

<u>conclusion</u>:-

assume that there is no significant difference between males and women reported they buy clothing from their mobile device

p1 = p2

c) <u>95% confidence intervals</u>

The confidence intervals are P± 1.96(√PQ/n)

we know that = p = \frac{n_{1}p_{1} +n_{2}p_{2} }{n_{1}+n_{2}}= \frac{400X0.59+300X0.52}{700}

after calculation we get P = 0.56 and Q =1-P =0.44

Confidence intervals are ( P- 1.96(√PQ/n), P+ 1.96(√PQ/n))

now substitute values , we get

( 0.56- 1.96(√0.56X0.44/700), 0.56+ 1.96(0.56X0.44/700))

on simplification we get (0.523 ,0.596)

Therefore the population proportion (0.56) lies in between the 95% of <u>confidence intervals  (0.523 ,0.596)</u>

<u></u>

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