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Paladinen [302]
3 years ago
9

An inheritance of $40,000 is divided among three investments yielding $3,500 in interest per year. The interest rates for the th

ree investments are 7%, 9%, and 11%. Find the amount of each investment when the second and third are $3,000 and $5,000 less than the first, respectively.
Mathematics
1 answer:
a_sh-v [17]3 years ago
4 0

Answer:

 x = $ 16,000  , y = $13,000  z= $11,000

Step-by-step explanation:

given,

Investment amount $40,000

three investments yielding = $3,500

three investments are 7%, 9%, and 11%

second and third are $3,000 and $5,000 less than the first  

        0.07 x + 0.09 y + 0.11 z = $ 3,500  

        x + (x - 3000 ) + (x - 5000) = $ 40,000

           3x = $48,000

             x = $ 16,000

     y = 16000 - 3000 = $ 13,000

     z = 16000 - 5000 = $ 11,000

the investment by each

 x = $ 16,000  , y = $13,000  z= $11,000

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A basketball player that shoots 80% from the free throw line attempts two free throws. The notation for conditional probability
Juliette [100K]

Answer:

P(made 2nd attempt|made 1st attempt)=P(made 2nd attempt)

Step-by-step explanation:

Here given that a basketball player that shoots 80% from the free throw line attempts two free throws.

If x is the no of shoots he makes (say) then we find that each throw is independent of the other.

In other words, because he made successful first attempt, his chances for second attempt will not change

Prob for success in each attempt remains the same as 0.80

Hence I throw is independent of II throw.

When A and B are independent,then we have

P(A/B) = P(A)

Hence answer is

P(made 2nd attempt|made 1st attempt)=P(made 2nd attempt)


8 0
3 years ago
WILL GIVE 20 POINTS
Rainbow [258]
The answer is D. 4^-2
5 0
3 years ago
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Which expression is a difference?
Lynna [10]
They added -2 + 1000 = 2
5 0
3 years ago
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Pleaseeee helpppppppppp
o-na [289]

Answer:

25 ft^2

Step-by-step explanation:

In direct variation, if y varies directly with x, then the equation has the form

y = kx,

where k is the constant of proportionality. y is proportional to x.

Let's call the area y and the distance x.

Here, the area varies with the square of the distance, so the equation has the form

y = kx^2

Here, y is proportional to the square of x.

We can find the value of k by using the given information.

y = kx^2

When x = 20 ft, y = 16 ft^2.

16 = k(20^2)

k = 16/400

k = 1/25

The equation of the relation is:

y = (1/25)x^2

Now we use the equation we found to answer the question.

What is y (the area) when x (the distance) is 25 ft?

y = (1/25)x^2

y = (1/25)(25^2)

y = 25

Answer: 25 ft^2

5 0
2 years ago
1) Use Newton's method with the specified initial approximation x1 to find x3, the third approximation to the root of the given
neonofarm [45]

Answer:

Check below, please

Step-by-step explanation:

Hello!

1) In the Newton Method, we'll stop our approximations till the value gets repeated. Like this

x_{1}=2\\x_{2}=2-\frac{f(2)}{f'(2)}=2.5\\x_{3}=2.5-\frac{f(2.5)}{f'(2.5)}\approx 2.4166\\x_{4}=2.4166-\frac{f(2.4166)}{f'(2.4166)}\approx 2.41421\\x_{5}=2.41421-\frac{f(2.41421)}{f'(2.41421)}\approx \mathbf{2.41421}

2)  Looking at the graph, let's pick -1.2 and 3.2 as our approximations since it is a quadratic function. Passing through theses points -1.2 and 3.2 there are tangent lines that can be traced, which are the starting point to get to the roots.

We can rewrite it as: x^2-2x-4=0

x_{1}=-1.1\\x_{2}=-1.1-\frac{f(-1.1)}{f'(-1.1)}=-1.24047\\x_{3}=-1.24047-\frac{f(1.24047)}{f'(1.24047)}\approx -1.23607\\x_{4}=-1.23607-\frac{f(-1.23607)}{f'(-1.23607)}\approx -1.23606\\x_{5}=-1.23606-\frac{f(-1.23606)}{f'(-1.23606)}\approx \mathbf{-1.23606}

As for

x_{1}=3.2\\x_{2}=3.2-\frac{f(3.2)}{f'(3.2)}=3.23636\\x_{3}=3.23636-\frac{f(3.23636)}{f'(3.23636)}\approx 3.23606\\x_{4}=3.23606-\frac{f(3.23606)}{f'(3.23606)}\approx \mathbf{3.23606}\\

3) Rewriting and calculating its derivative. Remember to do it, in radians.

5\cos(x)-x-1=0 \:and f'(x)=-5\sin(x)-1

x_{1}=1\\x_{2}=1-\frac{f(1)}{f'(1)}=1.13471\\x_{3}=1.13471-\frac{f(1.13471)}{f'(1.13471)}\approx 1.13060\\x_{4}=1.13060-\frac{f(1.13060)}{f'(1.13060)}\approx 1.13059\\x_{5}= 1.13059-\frac{f( 1.13059)}{f'( 1.13059)}\approx \mathbf{ 1.13059}

For the second root, let's try -1.5

x_{1}=-1.5\\x_{2}=-1.5-\frac{f(-1.5)}{f'(-1.5)}=-1.71409\\x_{3}=-1.71409-\frac{f(-1.71409)}{f'(-1.71409)}\approx -1.71410\\x_{4}=-1.71410-\frac{f(-1.71410)}{f'(-1.71410)}\approx \mathbf{-1.71410}\\

For x=-3.9, last root.

x_{1}=-3.9\\x_{2}=-3.9-\frac{f(-3.9)}{f'(-3.9)}=-4.06438\\x_{3}=-4.06438-\frac{f(-4.06438)}{f'(-4.06438)}\approx -4.05507\\x_{4}=-4.05507-\frac{f(-4.05507)}{f'(-4.05507)}\approx \mathbf{-4.05507}\\

5) In this case, let's make a little adjustment on the Newton formula to find critical numbers. Remember their relation with 1st and 2nd derivatives.

x_{n+1}=x_{n}-\frac{f'(n)}{f''(n)}

f(x)=x^6-x^4+3x^3-2x

\mathbf{f'(x)=6x^5-4x^3+9x^2-2}

\mathbf{f''(x)=30x^4-12x^2+18x}

For -1.2

x_{1}=-1.2\\x_{2}=-1.2-\frac{f'(-1.2)}{f''(-1.2)}=-1.32611\\x_{3}=-1.32611-\frac{f'(-1.32611)}{f''(-1.32611)}\approx -1.29575\\x_{4}=-1.29575-\frac{f'(-1.29575)}{f''(-4.05507)}\approx -1.29325\\x_{5}= -1.29325-\frac{f'( -1.29325)}{f''( -1.29325)}\approx  -1.29322\\x_{6}= -1.29322-\frac{f'( -1.29322)}{f''( -1.29322)}\approx  \mathbf{-1.29322}\\

For x=0.4

x_{1}=0.4\\x_{2}=0.4\frac{f'(0.4)}{f''(0.4)}=0.52476\\x_{3}=0.52476-\frac{f'(0.52476)}{f''(0.52476)}\approx 0.50823\\x_{4}=0.50823-\frac{f'(0.50823)}{f''(0.50823)}\approx 0.50785\\x_{5}= 0.50785-\frac{f'(0.50785)}{f''(0.50785)}\approx  \mathbf{0.50785}\\

and for x=-0.4

x_{1}=-0.4\\x_{2}=-0.4\frac{f'(-0.4)}{f''(-0.4)}=-0.44375\\x_{3}=-0.44375-\frac{f'(-0.44375)}{f''(-0.44375)}\approx -0.44173\\x_{4}=-0.44173-\frac{f'(-0.44173)}{f''(-0.44173)}\approx \mathbf{-0.44173}\\

These roots (in bold) are the critical numbers

3 0
2 years ago
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