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andre [41]
3 years ago
14

An electronics store received two shipments of radios. Of the 250 radios in the first shipment, exactly 6% were damaged. In the

second shipment, 4% of the radios were damaged. A total of 24 radios were damaged. How many radios in the combined shipments were NOT damaged? A. 466 B. 441 C. 451 D. 460
Mathematics
1 answer:
nataly862011 [7]3 years ago
7 0

Answer:

Total number of not damaged = 451

C. 451

Step-by-step explanation:

In the first shipment, 6% of the 250 radios where damaged.

Number of damaged= 6/100 * 250

Number of damaged= 6*2.5

Number of damaged= 15 radios.

Number of not damaged= 250-15

Number of not damaged = 235 radios

In the second shipment, 4% of the radios were damaged

Let the number of second shipment= y

0.04 of y where damaged

Total number of radio damaged= 24

15 + 0.04y = 24

0.04y= 24-15

0.04y = 9

Y= 9/0.04

Y= 225

Number of radio in the second shipment= 225

Number of damaged in the second shipment= 0.04*225

Number of damaged in the second shipment= 9

Number of not damaged in the second shipment= 225-9

Number of damaged in the second shipment= 216

Total number of not damaged

= 235+216

= 451

Total number of not damaged= 451

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A sample of eight workers in a clothing manufacturing company gave the following figures for the amount of time(in minutes) need
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Answer:

10.108 < \mu < 13.892    

Step-by-step explanation:

1) Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

\bar X represent the sample mean for the sample  

\mu population mean (variable of interest)

s represent the sample standard deviation

n represent the sample size  

We have the following distribution for the random variable:

X \sim N(\mu , \sigma=0.45)

And by the central theorem we know that the distribution for the sample mean is given by:

\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})

2) Confidence interval

The confidence interval for the mean is given by the following formula:

\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}   (1)

In order to calculate the mean and the sample deviation we can use the following formulas:  

\bar X= \sum_{i=1}^n \frac{x_i}{n} (2)  

s=\sqrt{\frac{\sum_{i=1}^n (x_i-\bar X)}{n-1}} (3)  

The mean calculated for this case is \bar X=12

The sample deviation calculated s=2.268

In order to calculate the critical value t_{\alpha/2} we need to find first the degrees of freedom, given by:

df=n-1=8-1=7

Since the Confidence is 0.95 or 95%, the value of \alpha=0.05 and \alpha/2 =0.025, and we can use excel, a calculator or a tabel to find the critical value. The excel command would be: "=-T.INV(0.025,7)".And we see that t_{\alpha/2}=2.36

Now we have everything in order to replace into formula (1):

12-2.36\frac{2.268}{\sqrt{8}}=10.108    

12+2.36\frac{2.268}{\sqrt{8}}=13.892

So on this case the 95% confidence interval would be given by (10.108;13.892)

10.108 < \mu < 13.892    

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