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Serga [27]
3 years ago
7

A statistical quality control process for cereal production measures the weight of a cereal box. The population standard deviati

on is known to be 0.09 ounces. If we wish to estimate the population average cereal box weight to within 0.011 ounces and be 94% confident, how large a sample should be used
Mathematics
1 answer:
algol133 years ago
5 0

Answer:

the sample should be 237

Step-by-step explanation:

The computation of the large the sample be used is shown below:

Given that

Standard deviation = 0.09 ounces

margin of error = 0.011

For 94% confidence interval, the z value be 1.88

As we know that

Margin of error = ((z score × standard deviation) ÷ (Sample size))^2

0.011 = (1.88 × 0.09) ÷ (Sample size))^2

0.011 = 0.1692 ÷ (Sample size))^2

So, the sample size be

= 15.38^2

= 236.60

Hence, the sample should be 237

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You start the month with $332.11 in your checking account. You have the following withdrawals during the month: $34.56, $223.00,
sergeinik [125]
In order to get the final value, you would just add and subtract the deposits/withdrawals from the initial amount in your account. 

$332.11 - $34.56 - $223.00 - $12.22 + 125.88 + 29.88 = $218.09

Your account balance is $218.09
5 0
3 years ago
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A nutrition laboratory tested 25 "reduced sodium" hotdogs of a certain brand, finding that the mean sodium content is 310 mg wit
inessss [21]

Answer:

The  95% confidence interval is  295.9 < \mu< 324.1

A   95% level of confidence mean that there is 95%  chance  that the true population mean will be in this interval

Step-by-step explanation:

From the question we are told that

    The sample size is  n  =  25

    The mean is  \= x  =  310 \ mg

     The standard deviation is  \sigma =  36 \ mg

Given that the confidence level is  95% then the level of significance is mathematically represented as

           \alpha  =  100 - 95

=>        \alpha  =  5\%

=>        \alpha  =  0.05

Next we obtain the critical value of  \frac{\alpha }{2} from the normal distribution table , the value is  

           Z_{\frac{\alpha }{2} } =Z_{\frac{0.05 }{2} }  =  1.96

Generally the margin of error is mathematically represented as

        E =  Z_{\frac{\alpha }{2} } *  \frac{\sigma }{\sqrt{n} }

substituting values

        E =  1.96 *  \frac{36 }{\sqrt{25} }

        E = 14.1

The 95% level of confidence interval  is mathematically represented as

      \= x - E < \mu

substituting values

     310- 14.1 < \mu< 310+ 14.1

     295.9 < \mu< 324.1

The  95% level of confidence mean that there is 95%  chance  that the true population mean will be in this interval

6 0
3 years ago
Need help right now !!!!! <br> Find the missing part
oksian1 [2.3K]

What kind of mathematics are you learning?

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3 years ago
3. Jim invests $3500, compounded annually at 12% for 3 years. What is the total
irga5000 [103]

Answer:

None of the above, it should be $4760

Step-by-step explanation:

First, converting R percent to r a decimal

r = R/100 = 12%/100 = 0.12 per year,

then, solving our equation

I = 3500 × 0.12 × 3 = 1260

I = $ 1,260.00

The simple interest accumulated

on a principal of $ 3,500.00

at a rate of 12% per year

for 3 years is $ 1,260.00

3 0
2 years ago
Find the equation of the line that is parallel to the line x + 5y = 10 and passes through the point (1, 3).
shusha [124]

Answer:

see explanation

Step-by-step explanation:

The equation of a line in slope- intercept form is

y = mx + c ( m is the slope and c the y- intercept )

Rearrange x + 5y = 10 into this form

Subtract x from both sides

5y = - x + 10 ( divide all terms by 5 )

y = - \frac{1}{5} x +2 ← in slope- intercept form

with slope m = - \frac{1}{5}

• Parallel lines have equal slopes, hence

y = - \frac{1}{5} x + c ← is the partial equation of the parallel line

To find c substitute (1, 3) into the partial equation

3 = - \frac{1}{5} + c ⇒ c = \frac{16}{5}

y = - \frac{1}{5} x + \frac{16}{5} ← equation of parallel line

3 0
2 years ago
Read 2 more answers
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