Answer:
b. 2.28%.
Step-by-step explanation:
Mean temperatue (μ) = 1000°F
Standard Deviation (σ) = 50
°F
For any temperature value, X, the z-score is given by:
For X= 900°F

A z-score of -2.0 corresponds to the 2.28-th percentile of a normal distribution. Therefore, the probability that X<900 is:

Answer:

Step-by-step explanation:
Since Quadrilateral ABCD ~ PQRS, therefore:

Let's find the value of x by using the ratio of two corresponding sides of both quadrilaterals. Let's use:

CD = 5
RS = 4.5
DA = 4
SP = x

Cross multiply


Divide both sides by 5


Answer:
E. equal to; Transitive Property
Step-by-step explanation: