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Gnesinka [82]
4 years ago
5

Need help please can someone help

Mathematics
1 answer:
Vladimir79 [104]4 years ago
8 0
Assuming this problem is a simplification problem, the best way to simplify a problem like this would be to combine the fractions. to combine fractions with different denominators, even with variables, would be to multiply and get equal denominators, and then simplify. To get equal fractions we simply multiply the 2nd denominator on both sides in the 1st and vice versa. In this case for the top we have 1/(z-4) and 2/(z+8), so then we multiply the denominators and get (z+8)/(z-4)(z+8) and 2(z-4)/(z-4)(z+8). To get the final numerator then we first multiply out the 2 and get 2z-8 on the right, then combine the two fractions by adding them, and get 3z/(z-4)(z+8). For the denominator we repeat the same process. We multiply the denominators to get 4(z-6)/(z+8)(z-6) and 3(z+8)/(z-6)(z+8) we simplify the numerators by multiplying the 4 and 3 to get 4z-24 and 3z+24, and then subtract. We then get 4z-24-3z-24 and so z-48 and (z-48)/(z-6)(z+8). So now we have 3z/(z-4)(z+8) for a numerator and (z-48)/(z-6)(z+8) as a denominator. we can multiply z-8 on both sides of the final fraction to cancel out the z-8, and so get 3z/(z-4)/(z-48)/(z-6). As you can no longer simplify anything this is your final fraction.
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A manufacturer of stone-ground deli-style mustard uses a high-speed machine to fill jars. The amount of mustard dispensed is nor
vladimir1956 [14]

This question is incomplete, the complete question is;

A manufacturer of stone-ground deli-style mustard uses a high-speed machine to fill jars. The amount of mustard dispensed is normally distributed with a mean weight of 290 grams, and a standard deviation of 4 grams. If the actual amount dispensed is too low, then their customers will be cheated; if it’s too high, then the company could lose money. To keep the machine properly calibrated, the company periodically takes a sample of 12 jars, to see if they need to stop production temporarily and re-calibrate. A recent sample produced a mean of 292.2 grams. Should the company be concerned that µ ≠ 290 grams? Use σ = 0:025. Use the p-value approach.

Answer:

P-Value = 0.08324

p-value ( 0.08324 ) is greater than significance level σ ( 0.025 );

fail to reject the null hypothesis

the company should not be concerned because, we have sufficient evidence to conclude that the mean weight is not different from 290 grams.

Step-by-step explanation:

Given the data in the question;

mean weight μ = 290 grams

sample mean x" = 292.2 grams

standard deviation s= 4 grams

sample size n = 12 jars

degree of freedom DF = n - 1 = 12 - 1 = 11

significance level σ  = 0.025

Two tailed Test

Null hypothesis             H₀ : μ = 290

Alternative hypothesis Hₐ : μ ≠ 290

Test Statistics;

t = ( x" - μ ) / ( s/√n )

we substitute our values into the equation

t = ( 292.2 - 290 ) / ( 4/√12 )

t = 2.2 / 1.1547

t = 1.905

From table; { t=1.905, df = 11, } Two tailed

P-Value = 0.08324

Hence, p-value ( 0.08324 ) is greater than significance level σ ( 0.025 );

fail to reject the null hypothesis meaning μ = 290

So the company should not be concerned because, we have sufficient evidence to conclude that the mean weight is not different from 290 grams.

3 0
3 years ago
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Rate 2 = ( 15 - 3 ) / 5

Rate 2 = 12 / 5

Rate 2 = 2.4 ft/ s

What is the rate of change between the 3rd second and the 5th second? That is, the change in speed. Which can be expressed as:

The rate of change = 5 - 2.4

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