Answer:
b) Julie can use properties of congruent triangles to show that AB≅BC and BC≅CD. Then she can show BC≅AD because opposite sides of a parallelogram are congruent
Step-by-step explanation:
On the assumption that Julie must show all four sides to be congruent, the one remaining step after using the properties of congruent triangles is to show that side AD is congruent to the rest of the sides. Answer choice B describes that.
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IMO, Julie is finished after she shows∠A≅∠B and AB≅BC, because a parallelogram will be a square if adjacent sides are congruent (makes it a rhombus) and adjacent angles are congruent (makes it a rectangle). A rhombus that is a rectangle is a square.
They are not parallel because the slopes are different (slope of 4 and slope of 0.25)
Neither are they perpendicular. If they were then m1*m2 ( the product of their slopes) would be -1 ). The product of these slopes = 4*0.25 = 1)
So choice A is the correct one.
Solution: We are given:
We need to find the z value corresponding to probability 0.84, in order to find the how much money almost 84% of gamblers spent at casino.
Using the standard normal table, we have:
Now we will use the z score formula to find the required amount:
approximately
Therefore, almost 84% of gamblers spent more than $720 amount of money at this casino.
Answer:
the value of x=
2×-12= -24
the value of y is 3y plus 20 which equals to 23
Answer:
a) 615
b) 715
c) 344
Step-by-step explanation:
According to the Question,
- Given that, A study conducted by the Center for Population Economics at the University of Chicago studied the birth weights of 732 babies born in New York. The mean weight was 3311 grams with a standard deviation of 860 grams
- Since the distribution is approximately bell-shaped, we can use the normal distribution and calculate the Z scores for each scenario.
Z = (x - mean)/standard deviation
Now,
For x = 4171, Z = (4171 - 3311)/860 = 1
- P(Z < 1) using Z table for areas for the standard normal distribution, you will get 0.8413.
Next, multiply that by the sample size of 732.
- Therefore 732(0.8413) = 615.8316, so approximately 615 will weigh less than 4171
- For part b, use the same method except x is now 1591.
Z = (1581 - 3311)/860 = -2
- P(Z > -2) , using the Z table is 1 - 0.0228 = 0.9772 . Now 732(0.9772) = 715.3104, so approximately 715 will weigh more than 1591.
- For part c, we now need to get two Z scores, one for 3311 and another for 5031.
Z1 = (3311 - 3311)/860 = 0
Z2 = (5031 - 3311)/860= 2
P(0 ≤ Z ≤ 2) = 0.9772 - 0.5000 = 0.4772
approximately 47% fall between 0 and 1 standard deviation, so take 0.47 times 732 ⇒ 732×0.47 = 344.
