He ate 1 - 6/8 which is 2/8, or 1/4 of an orange.
58% is about equal to 60%, 10% of 121 is 12.1.
12.1 * 6 = 72.6
To take this further, 58% is 2% less than 60%, or 2 times 1%. 1% of 121 is 1.21.
72.6 - (1.21 * 2) = 72.6 - 2.42 = 70.18
Answer:
Pr(X-Y ≤ 44.2) = 0.5593
Step-by-step explanation:
for a certain breed of terrier
Mean(μ) = 72cm
Standard deviation (σ) = 10cm
n = 64
For a certain breed of poodle
Mean(μ) = 28cm
Standard deviation (σ) = 5cm
n = 100
Let X be the random variable for the height of a certain breed of terrier
Let Y be the random variable for the height of a certain breed of poodle
μx - μy = 72 -28
= 44
σx - σy = √(σx^2/nx + σy^2/ny)
= √10^2/64 + 5^2/100
= √100/64 + 25/100
= √ 1.8125
= 1.346
Using normal distribution,
Z= (X-Y- μx-y) / σx-y
Z= (44.2 - 44) / 1.346
Z= 0.2/1.346
Z= 0.1486
From the Z table, Z = 0.149 = 0.0593
Φ(z) = 0 0593
The probability that the difference of the observed sample mean is at most 44.3 is Pr(Z ≤ 44.2)
Recall that if Z is positive,
Pr(Z≤a) = 0.5 + Φ(z)
Pr(Z ≤ 44.2) = 0.5 + 0.0593
= 0.5593
Therefore,
Pr(X-Y ≤ 44.2) = 0.5593
Let number of yellow marbles = x
Let number of red marbles = y
Let number of blue marbles = z
Given that Finley has 152 yellow, red, and blue marbles in a bag.
So we get equation:
x+y+z=152...(i)
"He has seven more red marbles than yellow marbles" means:
y=x+7...(ii)
"three times as many blue marbles as yellow marbles." means:
z=3x...(iii)
Now we just need to solve those equations to find the answer.
Plug (ii) and (iii) into (i)
x+(x+7)+(3x)=152
x+x+7+3x=152
Which looks similar to choice (B)
5x+7=152
5x=145
x=29
Hence final answer is choice B.
Required equation is
x+x+7+3x=152
Number of yellow marbles = 29
Given:
Vertices of triangle ABC are A (1,4), B(3,−2) and C(4,2).
Triangle ABC reflected over the x-axis to get the triangle A'B'C'.
To find:
The coordinates of the image A'B'C'.
Solution:
If a figure reflected over the x-axis, then rule of transformation is
Now, using this rule, we get
Therefore, the coordinates of the image A'B'C' after a reflection over the x-axis are A'(1,-4), B'(3,2) and C'(4,-2).
