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gulaghasi [49]
1 year ago
9

four friends contributed the sum of money to a charitable organization in the ratio 1:3:5:7. if the largest amount contrebuted i

s birr 35. calculate the total amount contributed by the four people​
Mathematics
1 answer:
suter [353]1 year ago
3 0

9514 1404 393

Answer:

  birr 80

Step-by-step explanation:

The total number of ratio units is 1+3+5+7 = 16. The largest amount is represented by 7 ratio units, so is 7/16 of the total. The total contributed is ...

  16/7 × (birr 35) = birr 80

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WARRIOR [948]

Answer:

We conclude that the new procedure will not decrease the population mean amount of time required to produce the part.

Step-by-step explanation:

We are given that a large manufacturing plant has analyzed the amount of time required to produce an electrical part and determined that the times follow a normal distribution with mean time μ = 45 hours.

A random sample of 25 parts will be selected and the average amount of time required to produce them will be determined. The sample mean amount of time is = 43.118 hours with the sample standard deviation s = 5.5 hours

<em>Let </em>\mu<em> = population mean amount of time required to produce an electrical part using new procedure</em>

SO, <u>Null Hypothesis</u>, H_0 : \mu \geq  45 hours   {means that the new procedure will remain same or increase the population mean amount of time required to produce the part}

<u>Alternate Hypothesis,</u> H_a : \mu < 45 hours   {means that the new procedure will decrease the population mean amount of time required to produce the part}

The test statistics that will be used here is <u>One-sample t test statistics </u>because we don't know about the population standard deviation;

              T.S.  = \frac{\bar X -\mu}{{\frac{s}{\sqrt{n} } } }  ~ t_n_-_1

where,  \mu = sample mean amount of time = 43.118 hours

             s = sample standard deviation = 5.5 hours

             n = sample of parts = 25

So, <u><em>test statistics</em></u>  =  \frac{43.118-45}{{\frac{5.5}{\sqrt{25} } } }  ~ t_2_4

                               =  -1.711

<em>Now at 0.025 significance level, the t table gives critical value of -2.064 at 24 degree of freedom for left-tailed test. Since our test statistics is more than the critical value of t so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region.</em>

Therefore, we conclude that the new procedure will remain same or increase the population mean amount of time required to produce the part.

4 0
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