The distribution is not a standard normal distribution.
We know that the variable (X) usually obeys and follows a standard normal distribution provided that the summation of (X) is zero and the variance of the random variable V(X) = 1.
Mathematically:
E(X) = 0 and V(X) = 1
Given that;
- the mean which follows a normal distribution (X) = 4.5338 g, and
- the standard deviation SD(X) = 0.1039 g
As such, the distribution is not a standard normal distribution.
Therefore, we can conclude that the distribution is not a standard normal distribution.
Learn more about standard normal distribution here:
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Answer:
(d) x = 3 or x = 4
Step-by-step explanation:
The zero product property can be used to find the solutions to a polynomial equation when it is written in factored form.
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Your polynomial can be put in factored form like this:
x² +12 = 7x . . . . . given
x² -7x +12 = 0 . . . . . standard form
(x -3)(x -4) = 0 . . . . . factored form*
x = 3 or x = 4 . . . . . . values of x that make the factors zero
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* The constants in the binomial factors must have a sum of -7 and a product of +12. It is often convenient to find them by listing factor pairs of 12:
12 = (-1)(-12) = (-2)(-6) = (-3)(-4)
Negative factors are used because we know the sum must be negative. Both have the same sign so their product is positive. The sums of the pairs shown are -13, -8, -7. The last pair is what we're looking for.
A=p (1+I/k^kn
A future value ?
P principle 3700
i interest rate 0.0184
K compounded quarterly 4
N time 6 years
A=3,700×(1+0.0184÷4)^(4×6)
A=4,130.84
Interest earned=A-p
4,130.84−3,700=430.84
Hope it helps:-)
Helps on what?
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