A factory produces 60005/12
1 answer:
Answer:
Step-by-step explanation:
Given the amount of tons of steel produced monthly by a factory to be 6000 5/12 tons, to determine their daily production, we need to use the conversion;
Maximum of 31 days = 1 month
This means that the steel company produces 6000 5/12 tons of steel in 31 days.
6000 5/12 tons = 31 dsays
72,005/12 tons = 31 days
in a day, the company will therefore produce approx. (72,005/12)/31 tons of steel.
(72,005/12)/31 = 72,005/12 * 1/31
= 72,005/12*31
= 72,005/372
= 193 209/372 tons of steel.
Hence, the factory produces about tons of steel daily
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option A i.e. Three-fourths (four-thirds) x = negative 6 (four-thirds).
<u>Step-by-step explanation:</u>
We have , The given expression as Three-fourths x = negative 6 , which can be written as . Now in order to solve this equation in one step , we must notice that coefficient of x must be 1 but it's
, Let's make coefficient of x as 1 by multiplying both side of equations by 4/3:
⇒
⇒
⇒
⇒
⇒
Therefore, x= -8 & correct option to solve the equation Three-fourths x = negative 6 for x in one step is <u>option A i.e. Three-fourths (four-thirds) x = negative 6 (four-thirds).</u>
Answer:
Step-by-step explanation:
Hello!
You have the variable
X: Area that can be painted with a can of spray paint (feet²)
The variable has a normal distribution with mean μ= 25 feet² and standard deviation δ= 3 feet²
since the variable has a normal distribution, you have to convert it to standard normal distribution to be able to use the tabulated accumulated probabilities.
a.
P(X>27)
First step is to standardize the value of X using Z= (X-μ)/ δ ~N(0;1)
P(Z>(27-25)/3)
P(Z>0.67)
Now that you have the corresponding Z value you can look for it in the table, but since tha table has probabilities of , you have to do the following conervertion:
P(Z>0.67)= 1 - P(Z≤0.67)= 1 - 0.74857= 0.25143
b.
There was a sample of 20 cans taken and you need to calculate the probability of painting on average an area of 540 feet².
The sample mean has the same distribution as the variable it is ariginated from, but it's variability is affected by the sample size, so it has a normal distribution with parameners:
X[bar]~N(μ;δ²/n)
So the Z you have to use to standardize the value of the sample mean is Z=(X[bar]-μ)/(δ/√n)~N(0;1)
To paint 540 feet² using 20 cans you have to paint around 540/20= 27 feet² per can.
c.
P(X≤27) = P(Z≤(27-25)/(3/√20))= P(Z≤2.98)= 0.999
d.
No. If the distribution is skewed and not normal, you cannot use the normal distribution to calculate the probabilities. You could use the central limit theorem to approximate the sampling distribution to normal if the sample size was 30 or grater but this is not the case.
I hope it helps!