I don't like dividing polynomials so I'm going to set x=1 and try them out.
(1-2-14+3)/4 = -3
Of the choices, only D gives -3 when x=1
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That's as much test-taking as math. If we were to divide
x^2 - 5x + 1
-------------------------------
x + 3 | x^3 - 2x^2 - 14x + 3
x^3 + 3x^2
-5x^2 - 14x
5x^2 + 15x
x + 3
x + 3
More work, still choice D
140 quarts of oil
49 quarts÷7 cars=7 quarts per car
7 quarts×20 cars=140 quarts
Answer:
a) 8.103 g
b) 9.2948
c) 0
Step-by-step explanation:
Given:
Data reported:
9.314 g, 9.215 g, 9.323 g, 8.103 g, 9.278 g, and 9.344 g
Now,
All the values except the 8.103 are above 9
Here the data 8.103 varies very much with respect to the other values
Hence,
a) the data 8.103 should be excluded
b) average value of the mass of the penny = 
= 9.2948 g
c) Deviation = Mean - Data
9.2948 - 9.314 = -0.0192
9.2948 - 9.215 = 0.0798
9.2948 - 9.323 = -0.0282
9.2948 - 9.278 = 0.0168
9.2948 - 9.344 = -0.0492
Thus,
Average deviation from mean = tex]\frac{-0.0192 + 0.0798 -0.0282 + 0.0168 -0.0492 }{5}[/tex]
= 0
Answer:
There are 118 plants that weight between 13 and 16 pounds
Step-by-step explanation:
For any normal random variable X with mean μ and standard deviation σ : X ~ Normal(μ, σ)
This can be translated into standard normal units by :
Let X be the weight of the plant
X ~ Normal( 15 , 1.75 )
To find : P( 13 < X < 16 )

= P( -1.142857 < Z < 0.5714286 )
= P( Z < 0.5714286 ) - P( Z < -1.142857 )
= 0.7161454 - 0.1265490
= 0.5895965
So, the probability that any one of the plants weights between 13 and 16 pounds is 0.5895965
Hence, The expected number of plants out of 200 that will weight between 13 and 16 = 0.5895965 × 200
= 117.9193
Therefore, There are 118 plants that weight between 13 and 16 pounds.
Answer:
1) m∠U = 90°
2) m∠C = 80°
Step-by-step explanation:
1) The given figure is a quadrilateral
The sum of the interior angles of quadrilateral = 360°
∴ The sum of the interior angles of the given figure = 360°
Therefore, we have;
80° + 24·x + 4 + 6 + 21·x + 90° = 360°
80° + 45·x + 10 + 90° = 360°
x = (360°- (80° + 10° + 90°))/45 = 4
x = 4
m∠U = 6 + 21·x = 6 + 21 × 4 = 90
m∠U = 90°
2) The sum of the interior angles of the given quadrilateral = 360°
∴ 21·x + 6 + 20·x + 24·x + 4 + 21·x + 6 = 360°
86·x + 16 = 360°
x = (360° - 16°)/86 = 4
x = 4
m∠C = 20·x = 20 × 4 = 80
m∠C = 80°
3) In the figure, some angles are left out, therefore, more information on the remaining angles required