Answer:
The amount needed to finance this level of withdrawals is $ 9422.06
Step-by-step explanation:
For a fixed amount periodically withdrawn, during periods, with an interest rate , there is a present value of:
. From where,
The amount needed to finance this level of withdrawals is $ 9422.0609
Answer:
y = -4x + 50
Step-by-step explanation:
To write the function we'll use y = mx + b, where y is the distance the boat is away from shore, m is how many km/minute the boat moves at, x is how many minutes it's been, and b is how many km away from shore the boat was at the start. We already know the value of b is 50, so we can put that into the equation:
y = mx + 50
To find how many km the boat moves per minute, let's use the given amount of minutes and km (9, 14) and put them into the equation:
14 = m(9) + 50
Now let's solve for m:
14 = 9m + 50
Subtract 50 from both sides to isolate the 9m:
14 - 50 = 9m + 50 - 50
- 36 = 9m
divide both sides by 9 to isolate the m:
-36/9 = 9m/9
-4 = m
The boat moves at -4 km/minute, which is the same thing as saying that after 1 minute passes, the boat gets 4km closer to the shore. Now we can input this value into the equation and we have our answer:
y = -4x + 50
Answer:
List all of the solutions.
50km=
9minutes=
14km=
Step-by-step explanation:
Answer:
14.
15.
Step-by-step explanation:
Solving (14):
First, we solve for x.
Since sides , then
This is so because the theorem of ASA (Angle-Side-Angle)
So, we have:
Collect Like Terms
Divide both sides by 3
Solving for m<B, m<C and m<D.
We simply substitute 11 for x in their respective expressions.
Solving (15):
First, we solve for x.
Since , then
This is so because the theorem of SAS (Side-Angle-Side)
So, we have:
Collect Like Terms
Divide both sides by 2
Solving for WX, WY and XY
We simply substitute 4 for x in their respective expressions.
Answer:
75
Step-by-step explanation:
Answer:
800 + 2400 = 3200
Step-by-step explanation:
Because 12 m = 1200 cm
A = L(2) + W(2)
A = 1200 (2) + 400 (2)
A = 2400 + 800
A = 3200
Answer:
a) The probability that both adults dine out more than once per week = 0.0253
b) The probability that neither adult dines out more than once per week = 0.7069
c) The probability that at least one of the two adults dines out more than once per week = 0.2931
d) Of the three events described, the event that can be considered unusual because of its low probability of occurring, 0.0253 (2.53%), is the event that the two randomly selected adults both dine out more than once per week.
Step-by-step explanation:
In a sample of 1200 U.S. adults, 191 dine out at a restaurant more than once per week.
Assuming this sample.is a random sample and is representative of the proportion of all U.S. adults, the probability of a randomly picked U.S. adult dining out at a restaurant more than once per week = (191/1200) = 0.1591666667 = 0.1592
Now, assuming this probability per person is independent of each other.
Two adults are picked at random from the entire population of U.S. adults, with no replacement, thereby making sure these two are picked at absolute random.
a) The probability that both adults dine out more than once per week.
Probability that adult A dines out more than once per week = P(A) = 0.1592
Probability that adult B dines out more than once per week = P(B) = 0.1592
Probability that adult A and adult B dine out more than once per week = P(A n B)
= P(A) × P(B) (since the probability for each person is independent of the other person)
= 0.1592 × 0.1592
= 0.02534464 = 0.0253 to 4 d.p.
b) The probability that neither adult dines out more than once per week.
Probability that adult A dines out more than once per week = P(A) = 0.1592
Probability that adult A does NOT dine out more than once per week = P(A') = 1 - P(A) = 1 - 0.1592 = 0.8408
Probability that adult B dines out more than once per week = P(B) = 0.1592
Probability that adult B does NOT dine out more than once per week = P(B') = 1 - P(B) = 1 - 0.1592 = 0.8408
Probability that neither adult dines out more than once per week = P(A' n B')
= P(A') × P(B')
= 0.8408 × 0.8408
= 0.70694464 = 0.7069 to 4 d.p.
c) The probability that at least one of the two adults dines out more than once per week.
Probability that adult A dines out more than once per week = P(A) = 0.1592
Probability that adult A does NOT dine out more than once per week = P(A') = 1 - P(A) = 1 - 0.1592 = 0.8408
Probability that adult B dines out more than once per week = P(B) = 0.1592
Probability that adult B does NOT dine out more than once per week = P(B') = 1 - P(B) = 1 - 0.1592 = 0.8408
The probability that at least one of the two adults dines out more than once per week
= P(A n B') + P(A' n B) + P(A n B)
= [P(A) × P(B')] + [P(A') × P(B)] + [P(A) × P(B)]
= (0.1592 × 0.8408) + (0.8408 × 0.1592) + (0.1592 × 0.1592)
= 0.13385536 + 0.13385536 + 0.02534464
= 0.29305536 = 0.2931 to 4 d.p.
d) Which of the events can be considered unusual? Explain.
The event that can be considered as unusual is the event that has very low probabilities of occurring, probabilities of values less than 5% (0.05).
And of the three events described, the event that can be considered unusual because of its low probability of occurring, 0.0253 (2.53%), is the event that the two randomly selected adults both dine out more than once per week.
Hope this Helps!!!