I will mark Brainliest! 6th grade math work
2 answers:
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<h3>Answer:</h3>
- <u>20</u> kg of 20%
- <u>80</u> kg of 60%
I like to use a little X diagram to work mixture problems like this. The constituent concentrations are on the left; the desired mix is in the middle, and the right legs of the X show the differences along the diagonal. These are the ratio numbers for the constituents. Reducing the ratio 32:8 gives 4:1, which totals 5 "ratio units". We need a total of 100 kg of alloy, so each "ratio unit" stands for 100 kg/5 = 20 kg of constituent.
That is, we need 80 kg of 60% alloy and 20 kg of 20% alloy for the product.
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<em>Using an equation</em>
If you want to write an equation for the amount of contributing alloy, it works best to let a variable represent the quantity of the highest-concentration contributor, the 60% alloy. Using x for the quantity of that (in kg), the amount of copper in the final alloy is ...
... 0.60x + 0.20(100 -x) = 0.52·100
... 0.40x = 32 . . . . . . . . . . .collect terms, subtract 20
... x = 32/0.40 = 80 . . . . . kg of 60% alloy
... (100 -80) = 20 . . . . . . . .kg of 20% alloy
Answer:
Step-by-step explanation:
If we assume that the number of arrivals is normally distributed and we don't know the population standard deviation, we can calculated a 95% confidence interval to estimate the mean value as:
where x' is the population mean value, x is the sample mean value, s is the sample standard deviation, n is the size of the sample, is equal to 0.05 (it is calculated as: 1 - 0.95) and
is the t value with n-1 degrees of freedom that let a probability of
on the right tail.
So, replacing the mean of the sample by 49, the standard deviation of the sample by 17.38, n by 10 and by 2.2621 we get:
Finally, the interval values that she get is: