A. z = 0.74
The z-score of 0.74 translates to a percentile of 0.77035. Hence, the area under the standard normal curve to the left of z-score 0.74 is ~0.77.
b. z = -2.16
This z-score translates to a percentile of 0.015386 which is also the numerical value of the area under the curve to the left of the z-score
c. z = 1.02
The percentile equivalent of the z-score above is 0.846. The area is also 0.846.
d. z = -0.15
The percentile equivalent and the area is equal to 0.44.
Answer:
Step-by-step explanation:
This is a unit rate problem. If we want to find out how much chocolate was eaten in 1 minute, we divide 31.5 by 7. This gives us the number of ounces of chocolate eaten in 1 minute: 4.5 ounces per minute.
The correct answer is C
Hope this helped :)
<h3>
Answer: 133</h3>
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Explanation:
The quickest way to get this answer is to add the angles given to get 87+46 = 133
This is through the use of the remote interior angle theorem.
Note how the angles 87 and 46 are interior, or inside the triangle. And also, they are not adjacent to the exterior angle we want to find. So that's where the "remote" portion comes in.
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The slightly longer method involves letting x be the measure of the missing interior angle of the triangle.
The three interior angles add to 180
87+46+x = 180
133+x = 180
x = 180 - 133
x = 47
The missing interior angle of the triangle is 47 degrees.
Angle 1 is adjacent and supplementary to this 47 degree angle, so,
(angle1)+(47) = 180
angle1 = 180-47
angle1 = 133 degrees
This example helps confirm that the remote interior angle theorem is correct.
Answer:
Given:
In Rhombus QRST, diagonals QS and RT intersect at W and U∈QR and point V∈RT such that UV⊥QR. (shown in below diagram)
To prove: QW•UR =WT•UV
Proof:
In a rhombus diagonals bisect perpendicularly,
Thus, in QRST
QW≅WS, WR ≅ WT and m∠QWR=m∠QWT=m∠RWS=m∠TWS=90°.
In triangles QWR and UVR,
(Right angles)
(Common angles)
By AA similarity postulate,
The corresponding sides in similar triangles are in same proportion,
(∵ WR ≅ WT )
Hence, proved.
