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2 years ago

5

A person invested $7,200 in an account growing rate allowing money to double every 8 years. How much money would be in the accou

nt after 19 years, to the nearest dollar?

1 answer:

spayn [35]2 years ago

5 0

273600 in his bank account

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find the area of a rectangle that has a length of 8.3 inches. and a width that is 0.6 times the length. show work please

grigory [225]

Find the width first...8.3×0.6.....then find the area....length × width= 8.3×4.9=40.67....that's what I got

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3 years ago

Read 2 more answers

An arcade booth at a county fair has a person pick a coin from two possible coins available and then toss it. If the coin chosen

nydimaria [60]

Answer:

<em>b. 0.6024</em>

Step-by-step explanation:

<u>Conditional Probability</u>

Suppose two events A and B are not independent, i.e. they can occur simultaneously. It means there is a space where the intersection of A and B is not empty:

$P(A\cap B) \neq 0$

If we already know event B has occurred, we can compute the probability that event A has also occurred with the conditional probability formula

$\displaystyle P(A|B)=\frac{P(A\cap B)}{P(B)}$

Now analyze the situation presented in the question. Let's call F to the fair coin with 50%-50% probability to get heads-tails, and U to the unfair coin with 32%-68% to get heads-tails respectively.

Since the probability to pick either coin is one half each, we have

$P(U)=P(F)=50\%=0.5$

If we had picked the fair coin, the probability of getting heads is 0.5 also, so

$P(F\cap H)=0.5\cdot 0.5=0.25$

If we had picked the unfair coin, the probability of getting heads is 0.32, so

$P(U\cap H)=0.32\cdot 0.5=0.16$

Being A the event of choosing the fair coin, and B the event of getting heads, then

$P(B)=P(F\cap H)+P(U\cap H)=0.25+0.16=0.41$

$P(A\cap B)=P(F\cap H)=0.25$

$\displaystyle P(A|B)=\frac{0.25}{0.41}=0.61$

The closest answer is

b. 0.6024

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3 years ago

Lily begins solving the equation 4(x-1)-x=3(x+5)-11. Her work is shown below. 4(x-1) - x = 3 (x+5)-11 4x −4) −x =3x +15 −11 3x-4

aivan3 [116]

Answer:

Um...

Step-by-step explanation:

Hai

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2 years ago

Find the standard form of the equation of the ellipse with the given characteristics. Vertices: (4, 0), (4, 14); endpoints of th

saul85 [17]

Answer:

The standard form of the ellipse is $\frac{(x-4)^{2}}{9} + \frac{(y-7)^{2}}{49} = 1$ .

Step-by-step explanation:

The major axis of the ellipse is located in the y axis, whereas the minor axis is in the x axis. The center of the ellipse is the midpoint of the line segment between vertices, this is:

$(h, k) =\frac{1}{2}\cdot V_{1} (x,y) + \frac{1}{2}\cdot V_{2} (x,y)$ (1)

If we know that $V_{1} (x,y) = (4,0)$ and $V_{2}(x,y) = (4, 14)$ , then the coordinates of the center are, respectively:

$(h,k) = \frac{1}{2}\cdot (4, 0) + \frac{1}{2}\cdot (4,14)$

$(h,k) = (2,0) + (2, 7)$

$(h, k) = (4, 7)$

The length of each semiaxis is, respectively:

$a = \sqrt{(1 - 4)^{2}+(7-7)^{2}}$

$a = 3$

$b = \sqrt{(4-4)^{2}+(0-7)^{2}}$

$b = 7$

The standard equation of the ellipse is described by the following formula:

$\frac{(x-h)^{2}}{a^{2}}+ \frac{(y-k)^{2}}{b^{2}} = 1$

Where:

$h$ , $k$ - Coordinates of the center of the ellipse.

$a$ , $b$ - Length of the orthogonal semiaxes.

If we know that $h = 4$ , $k = 7$ , $a = 3$ and $b = 7$ , then the standard form of the ellipse is:

$\frac{(x-4)^{2}}{9} + \frac{(y-7)^{2}}{49} = 1$

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3 years ago

Please help. Will reward brainliest

ipn [44]

Answer:

c is the answer

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2 years ago