randy deposited $1000 into an account that paid 2.8% with continuous compounding. what was her balance after 6 years

Find an equation of the tangent line to the curve at the given point. y = x + tan(x) at (pi, pi)

S_A_V [24]

<span>Differentiating :

y' = 1 + sec^2(x),
</span><span>
cosπ=−1</span><span>

plug in pi = -1 ,for x:

1 + sec^2(pi) = 1 + (-1)^2
= 2</span>

8 0

3 years ago

Suppose the function f(t) = 95 cosine (StartFraction pi Over 10 EndFraction t) + 120 models the height of a seat on a Ferris whe

drek231 [11]

Answer:

im not 100% bit i think its 95ft

Step-by-step explanation:

5 0

3 years ago

Read 2 more answers

What time is 10 minutes before 8:00 in the morning? *

strojnjashka [21]

Answer: 7:50 AM

Step-by-step explanation:

The clock goes to 60 minutes so if you subtract 10 from 8:00 AM

It will start at 7:60

7:60-10 is 7:50

6 0

3 years ago

Read 2 more answers

Both families use 1250 kWh for the given 30 day period. The Freeman family uses 400 kWh during on-peak hours and 850 during off-

Natalka [10]

Ok ummm I honestly don’t know I’m so sorry I couldn’t help you wit this prob

8 0

3 years ago

The prior probabilities for events A1 and A2 are P(A1) = 0.20 and P(A2) = 0.80. It is also known that P(A1 ∩ A2) = 0. Suppose P(

Umnica [9.8K]

Answer:

(a) $A_1$ and $A_2$ are indeed mutually-exclusive.

(b) $\displaystyle P(A_1\; \cap \; B) = \frac{1}{20}$ , whereas $\displaystyle P(A_2\; \cap \; B) = \frac{1}{25}$ .

(c) $\displaystyle P(B) = \frac{9}{100}$ .

(d) $\displaystyle P(A_1 \; |\; B) \approx \frac{5}{9}$ , whereas $P(A_1 \; |\; B) = \displaystyle \frac{4}{9}$

Step-by-step explanation:

$P(A_1 \; \cap \; A_2) = 0$ means that it is impossible for events $A_1$ and $A_2$ to happen at the same time. Therefore, event $A_1$ and $A_2$ are mutually-exclusive.

By the definition of conditional probability:

$\displaystyle P(B \; | \; A_1) = \frac{P(B \; \cap \; A_1)}{P(B)} = \frac{P(A_1 \; \cap \; B)}{P(B)}$ .

Rearrange to obtain:

$\displaystyle P(A_1 \; \cap \; B) = P(B \; |\; A_1) \cdot P(A_1) = 0.25 \times 0.20 = \frac{1}{20}$ .

Similarly:

$\displaystyle P(A_2 \; \cap \; B) = P(B \; |\; A_2) \cdot P(A_2) = 0.80 \times 0.05 = \frac{1}{25}$ .

Note that:

$\begin{aligned}P(A_1 \; \cup \; A_2) &= P(A_1) + P(A_2) - P(A_1 \; \cap \; A_2) = 0.20 + 0.80 = 1\end{aligned}$ .

In other words, $A_1$ and $A_2$ are collectively-exhaustive. Since $A_1$ and $A_2$ are collectively-exhaustive and mutually-exclusive at the same time:

$\displaystyle P(B) = P(B \; \cap \; A_1) + P(B \; \cap \; A_2) = \frac{1}{20} + \frac{1}{25} = \frac{9}{100}$ .

By Bayes' Theorem:

$\begin{aligned} P(A_1 \; |\; B) &= \frac{P(B \; | \; A_1) \cdot P(A_1)}{P(B)} \\ &= \frac{0.25 \times 0.20}{9/100} = \frac{0.05 \times 100}{9} = \frac{5}{9}\end{aligned}$ .

Similarly:

$\begin{aligned} P(A_2 \; |\; B) &= \frac{P(B \; | \; A_2) \cdot P(A_2)}{P(B)} \\ &= \frac{0.05 \times 0.80}{9/100} = \frac{0.04 \times 100}{9} = \frac{4}{9}\end{aligned}$ .

6 0

2 years ago