A used car is 34,000
You take 40,000 and multiply it by 0.85
56 because a quartile is a type of quantile. The first quartile (Q 1) is defined as the middle number between the smallest number and the median of the data set.
Answer:
![(7x)^{\frac{2}{3} = (\sqrt[3]{7x})^2](https://tex.z-dn.net/?f=%20%287x%29%5E%7B%5Cfrac%7B2%7D%7B3%7D%20%3D%20%28%5Csqrt%5B3%5D%7B7x%7D%29%5E2%20)
Step-by-step explanation:
Given the expression
, to express this as a radical expressions, we'd apply the rule/law of indices that deals with converting expressions that has rational exponents into radical expressions.
The rule of indices to apply is: ![b^{\frac{m}{n}} = (\sqrt[n]{b})^m](https://tex.z-dn.net/?f=%20b%5E%7B%5Cfrac%7Bm%7D%7Bn%7D%7D%20%3D%20%28%5Csqrt%5Bn%5D%7Bb%7D%29%5Em%20)
To apply this to the expression,
, the denominator of the fraction of the exponent would determine the root, that is, cube root in this case. The numerator of the exponent would then determine the exponent of the radical expressions.
Thus:
![(7x)^{\frac{2}{3} = (\sqrt[3]{7x})^2](https://tex.z-dn.net/?f=%20%287x%29%5E%7B%5Cfrac%7B2%7D%7B3%7D%20%3D%20%28%5Csqrt%5B3%5D%7B7x%7D%29%5E2%20)
Isn’t the same ?? so wouldn’t it be 100 also ?
Answer:
72
Step-by-step explanation:
Let
Carl grades = x = 62, 78, 59, 89
Number of grades, N = 4
Mean of Carl's grade = sum of x / number of grades,N
= (62 + 78 + 59 + 89) / 4
= 288/4
= 72
Therefore, mean of Carl's grade = 72