Answer:
-68 is your answer
Step-by-step explanation:
Remember to follow PEMDAS (Parenthesis, Exponents, Multiplication, Division, Addition, Subtraction)
First, solve each parenthesis
(-21 x 3) = -63
(15/-3) = -5
Next, add
(-63) + (-5) = -63 - 5 = -68
-68 is your answer
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What is the full question. I would be able to help if I saw it all.
Answer:
C. DC = DC. By reflexive property of equality.
Step-by-step explanation:
We are given that,
In step 4, Jimmy proved ΔADC ≅ ΔBCD by the SAS Congruence Postulate.
For that, the previous steps states,
Step 1: AD = BC, as opposite sides of a rectangle are congruent.
Step 2: ∠ADC = ∠BCD
So, in order to satisfy the SAS Postulate, we must have another pair of sides equal from the respective rectangles.
As, we can see that in ΔADC and ΔBCD, the side DC is a common side.
Thus, in the missing step, we get,
Step 3: DC = DC, By the reflexive property of equality.
Hence, option C is correct.
<u>Question 1 solution:</u>
You have two unknowns here:
Let the Water current speed = W
Let Rita's average speed = R
We are given <em>two </em>situations, where we can form <em>two equations</em>, and therefore solve for the <em>two unknowns, W, R</em>:
Part 1) W→ , R←(against current, upstream)
If Rita is paddling at 2mi/hr against the current, this means that the current is trying to slow her down. If you look at the direction of the water, it is "opposing" Rita, it is "opposite", therefore, our equation must have a negative sign for water<span>:
</span>R–W=2 - equation 1
Part 2) W→ , R<span>→</span>(with current)
Therefore, R+W=3 - equation 2
From equation 1, W=R-2,
Substitute into equation 2.
R+(R–2)=3
2R=5
R=5/2mi/hr
So when W=0 (still), R=5/2mi/hr
Finding the water speed using the same rearranging and substituting process:
1... R=2+W
2... (2+W)+W=3
2W=1
W=1/2mi/hr
(a) Average time to get to school
Average time (minutes) = Summation of the two means = mean time to walk to bus stop + mean time for the bust to get to school = 8+20 = 28 minutes
(b) Standard deviation of the whole trip to school
Standard deviation for the whole trip = Sqrt (Summation of variances)
Variance = Standard deviation ^2
Therefore,
Standard deviation for the whole trip = Sqrt (2^2+4^2) = Sqrt (20) = 4.47 minutes
(c) Probability that it will take more than 30 minutes to get to school
P(x>30) = 1-P(x=30)
Z(x=30) = (mean-30)/SD = (28-30)/4.47 ≈ -0.45
Now, P(x=30) = P(Z=-0.45) = 0.3264
Therefore,
P(X>30) = 1-P(X=30) = 1-0.3264 = 0.6736 = 67.36%
With actual average time to walk to the bus stop being 10 minutes;
(d) Average time to get to school
Actual average time to get to school = 10+20 = 30 minutes
(e) Standard deviation to get to school
Actual standard deviation = Previous standard deviation = 4.47 minutes. This is due to the fact that there are no changes with individual standard deviations.
(f) Probability that it will take more than 30 minutes to get to school
Z(x=30) = (mean - 30)/Sd = (30-30)/4.47 = 0/4.47 = 0
From Z table, P(x=30) = 0.5
And therefore, P(x>30) = 1- P(X=30) = 1- P(Z=0.0) = 1-0.5 = 0.5 = 50%
