Answer:
Its B, y=-0.89x^2+3.24x+11.93
Step-by-step explanation:
mark brainest pls
Answer:
x=6. x=15
Step-by-step explanation:
13. x=6
16. x=15
For Data Set B, we see that the data is more varied. The absolute deviations are 4, 3, 2, 5. The average of these absolute deviations is 3.5. MAD_B = (4+3+2+5)/4 =3.5 M ADB
Hence, The average of these absolute deviations is 3.5.
Answer:
a = 3, b = 0, c = 0, d = -2
Step-by-step explanation:
<em>To find the reflection Multiply the matrices</em>
∵ The dimension of the first matrix is 2 × 2
∵ The dimension of the second matrix is 2 × 3
<em>1. Multiply the first row of the 1st matrix by each column in the second matrix add the products of each column to get the first row in the 3rd matrix.</em>
2. Multiply the second row of the 1st matrix by each column in the second matrix add the products of each column to get the second row of the 3rd matrix
×
= ![\left[\begin{array}{ccc}(1*0+0*0)&(1*3+0*0)&(1*0+0*2)\\(0*0+-1*0)&(0*3+-1*0)&(0*0+-1*2)\end{array}\right]=\left[\begin{array}{ccc}0&3&0\\0&0&-2\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D%281%2A0%2B0%2A0%29%26%281%2A3%2B0%2A0%29%26%281%2A0%2B0%2A2%29%5C%5C%280%2A0%2B-1%2A0%29%26%280%2A3%2B-1%2A0%29%26%280%2A0%2B-1%2A2%29%5Cend%7Barray%7D%5Cright%5D%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D0%263%260%5C%5C0%260%26-2%5Cend%7Barray%7D%5Cright%5D)
Compare the elements in the answer with the third matrix to find the values of a, b, c, and d
∴ a = 3
∴ b = 0
∴ c = 0
∴ d = -2
The following is the rest of the question:
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Mikey's results showed that although both mice gained weight over the
month, mouse 2 gained more weight than mouse 1.Which graph below best
shows these results?
The complete figure is attached
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Solution:
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Graph (1)⇒⇒⇒ mouse 2 has constant weight and mouse 1 gained weight
Graph (3)⇒⇒⇒ mouse 2 has did not gain weight and mouse 1 gained weight
∴ Graph (1) and (3) are incorrect because both mice gained weight .
Graph (4) is incorrect because mouse 2 gained more weight than mouse 1
So, the correct answer is graph (2)