Answer:
so 1- 2 is negative 3 so minus tha by 3 than 4x2
Step-by-step explanation:
(-2,7)(3,-3)
slope(m) = (-3-7) / (3 - (-2) = -10/(3 + 2) = -10/5 = -2
y = mx + b
slope(m) = -2
use either of ur points (-2,7)...x = -2 and y = 7
now sub and find b, the y int
7 = -2(-2) + b
7 = 4 + b
7 - 4 = b
3 = b
so ur equation is : y = -2x + 3 or 2x + y = 3
I don't know.. what happens next?
Answer:
2.3 × 10^-7
Step-by-step explanation:
First, create an equation with the initial decimal being the coefficient multiplied by 10 to the power of 0.
(For example, let’s create the starting scientific notation equation for the decimal 475,000).
475,000 = 475,000 × 10^0
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Now that we have an equation, the second step is to move the decimal point in the coefficient until there is 1 significant digit to the left of the decimal point. For each space the decimal point is moved to the left, increase the exponent by 1.
(Continuing the example above, move the decimal point to the left and increment the exponent.).
475,000 = 475,000.0 × 10^0
475,000 = 47,500.0 × 10^1
475,000 = 4,750.0 × 10^2
475,000 = 475.0 × 10^3
475,000 = 47.5 × 10^4
475,000 = 4.75 × 10^5
Answer:
It is more reasonable to model the standard class with normal distribution
Step-by-step explanation:
- Standard Class:
The range of grades/marks obtained seems normally distributed with a reasonable average of 65% and covering the entire spectrum. Since most of the students score around average 65% the bump of normal bell curve is well situated about the mean.
- Advanced Class:
The range of grades/marks obtained seems concentrated at around 2 means one for the students with higher ability of mathematics and one for the students with lower ability of mathematics ( Higher for english ). The mathematics abled score in the upper percentile while the English abled score in the mid-lower percentile. The deviation of average scores for is large enough to make this a bimodal distribution rather than a normal distribution. However, there's strong correlation between excelling in one academic subject and excelling in another.
Conclusion:
It is more reasonable to model the standard class with normal distribution