Answer:
B
as [tex]\cos(\theta)=\srqt{1-\sin^2(\theta)}\[tex]
and since theta is in second quadrant, where cosine is negative,
so -√(1-z²)
The value of y will be 18 or 144.
Given information:
The given expression is ![\sqrt[3]{\frac{144}{y} }](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B%5Cfrac%7B144%7D%7By%7D%20%7D)
.
It is required to find the values of y which are whole numbers.
Now, factorize 144 as,
So, for the value of given expression to be a whole number, the value of y should be,
or 144.
For the above values of y, the given expression will be,
![\sqrt[3]{\frac{144}{y} }=\sqrt[3]{\frac{144}{18} }\\=\sqrt[3]{8} =2\\\sqrt[3]{\frac{144}{y} }=\sqrt[3]{\frac{144}{144} }\\=\sqrt[3]{1} \\=1](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B%5Cfrac%7B144%7D%7By%7D%20%7D%3D%5Csqrt%5B3%5D%7B%5Cfrac%7B144%7D%7B18%7D%20%7D%5C%5C%3D%5Csqrt%5B3%5D%7B8%7D%20%3D2%5C%5C%5Csqrt%5B3%5D%7B%5Cfrac%7B144%7D%7By%7D%20%7D%3D%5Csqrt%5B3%5D%7B%5Cfrac%7B144%7D%7B144%7D%20%7D%5C%5C%3D%5Csqrt%5B3%5D%7B1%7D%20%5C%5C%3D1)
Therefore, the value of y will be 18 or 144.
For more details, refer to the link:
brainly.com/question/17429689
It is possible to calculate mathematically the area under the normal curve between any two values of z.
However, tables/software have been developed to give the areas under the normal curve to the left of particular values of z. The function is the probability of Z<z, or P(Z<z).
The area between two values z1 and z2 (where z2>z1) is therefore
P(Z<z2)-P(Z<z1).
For example, to find the area between z1=1.5, z2=2.5
is
P(Z<2.5)-P(Z<1.5)
=0.99379-0.93319
=0.06060
(above values obtained by software, such as R)
For example, the value P(Z<2.5) can be calculated using
P(Z<2.5)=erf(2.5/sqrt(2))/2+1/2
where erf(x) is a mathematical function that does not have an explicit formula (calculated by summation of series, or tabulated).
You had $9.28 prior to making the withdrawal
-$20.51 = x - $29.79
x = $9.28
