Let us model this problem with a polynomial function.
Let x = day number (1,2,3,4, ...)
Let y = number of creatures colled on day x.
Because we have 5 data points, we shall use a 4th order polynomial of the form
y = a₁x⁴ + a₂x³ + a₃x² + a₄x + a₅
Substitute x=1,2, ..., 5 into y(x) to obtain the matrix equation
| 1 1 1 1 1 | | a₁ | | 42 |
| 2⁴ 2³ 2² 2¹ 2⁰ | | a₂ | | 26 |
| 3⁴ 3³ 3² 3¹ 3⁰ | | a₃ | = | 61 |
| 4⁴ 4³ 4² 4¹ 4⁰ | | a₄ | | 65 |
| 5⁴ 5³ 5² 5¹ 5⁰ | | a₅ | | 56 |
When this matrix equation is solved in the calculator, we obtain
a₁ = 4.1667
a₂ = -55.3333
a₃ = 253.3333
a₄ = -451.1667
a₅ = 291.0000
Test the solution.
y(1) = 42
y(2) = 26
y(3) = 61
y(4) = 65
y(5) = 56
The average for 5 days is (42+26+61+65+56)/5 = 50.
If Kathy collected 53 creatures instead of 56 on day 5, the average becomes
(42+26+61+65+53)/5 = 49.4.
Now predict values for days 5,7,8.
y(6) = 152
y(7) = 571
y(8) = 1631
Answer:
A
Step-by-step explanation:
The two angles in the triangle in the question are 58 degrees and 54 degrees.
Three angles in a triangle add up to 180 degrees
58 + 54 + x = 180
112 + x = 180
subtract 112 on both sides
x = 180 - 112 = 68 degrees
Only option A has a triangle that has the same angles, thus triangles are similar.
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The measure of center and range to be used to compare the population is the median and the inter quartile range.
<u>Explanation:</u>
You utilize the median and the interquartile range (IQR) to portray skewed distributions of information. To look at two populaces, utilize the mean and the MAD when the two populations are symmetric. Utilize the middle and the IQR when possibly one or the two circulations are slanted.
In statistics and probability theory, the median is the worth isolating the higher half from the lower half of an information test, a populace or a likelihood circulation. For an informational collection, it might be thought of as the "center" esteem.
Answer:
The time taken for the projectile to reach the given height is 1.2 s.
Step-by-step explanation:
Given;
height of travel, h = 25 ft
initial velocity of the projectile, u = 15 ft/s
The time taken for the projectile to travel a height of 25 ft is given by the following kinematic equation;
h = ut + ¹/₂gt²
25 = 15t + ¹/₂(9.8)t²
25 = 15t + 4.9t²
4.9t² + 15t - 25 = 0
solving the quadratic equation, we will have the following solution of t;
t = 1.2 s
Therefore, the time taken for the projectile to reach the given height is 1.2 s.
