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3 years ago

7

going into the final exam, which will count as two tests, amelia has test scores of 77 84 70 59 and 92. what score does amelia n

eed on the final in order to have an average score of 80?

1 answer:

Fudgin [204]3 years ago

6 0

Answer: 85 is your answer

Step-by-step explanation:

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<img src="https://tex.z-dn.net/?f=%24a%2Ba%20r%2Ba%20r%5E%7B2%7D%2B%5Cldots%20%5Cinfty%3D15%24%24a%5E%7B2%7D%2B%28a%20r%29%5E%7B

riadik2000 [5.3K]

Let

$S_n = \displaystyle \sum_{k=0}^n r^k = 1 + r + r^2 + \cdots + r^n$

where we assume |r| < 1. Multiplying on both sides by r gives

$r S_n = \displaystyle \sum_{k=0}^n r^{k+1} = r + r^2 + r^3 + \cdots + r^{n+1}$

and subtracting this from $S_n$ gives

$(1 - r) S_n = 1 - r^{n+1} \implies S_n = \dfrac{1 - r^{n+1}}{1 - r}$

As n → ∞, the exponential term will converge to 0, and the partial sums $S_n$ will converge to

$\displaystyle \lim_{n\to\infty} S_n = \dfrac1{1-r}$

Now, we're given

$a + ar + ar^2 + \cdots = 15 \implies 1 + r + r^2 + \cdots = \dfrac{15}a$

$a^2 + a^2r^2 + a^2r^4 + \cdots = 150 \implies 1 + r^2 + r^4 + \cdots = \dfrac{150}{a^2}$

We must have |r| < 1 since both sums converge, so

$\dfrac{15}a = \dfrac1{1-r}$

$\dfrac{150}{a^2} = \dfrac1{1-r^2}$

Solving for r by substitution, we have

$\dfrac{15}a = \dfrac1{1-r} \implies a = 15(1-r)$

$\dfrac{150}{225(1-r)^2} = \dfrac1{1-r^2}$

Recalling the difference of squares identity, we have

$\dfrac2{3(1-r)^2} = \dfrac1{(1-r)(1+r)}$

We've already confirmed r ≠ 1, so we can simplify this to

$\dfrac2{3(1-r)} = \dfrac1{1+r} \implies \dfrac{1-r}{1+r} = \dfrac23 \implies r = \dfrac15$

It follows that

$\dfrac a{1-r} = \dfrac a{1-\frac15} = 15 \implies a = 12$

and so the sum we want is

$ar^3 + ar^4 + ar^6 + \cdots = 15 - a - ar - ar^2 = \boxed{\dfrac3{25}}$

which doesn't appear to be either of the given answer choices. Are you sure there isn't a typo somewhere?

7 0

2 years ago

The figure here shows triangle AOC inscribed in the region cut from the parabola y=x^2 by the line y=a^2. Find the limit of the

aleksandrvk [35]

Area of the parabolic region = Integral of [a^2 - x^2 ]dx | from - a to a =

(a^2)x - (x^3)/3 | from - a to a = (a^2)(a) - (a^3)/3 - (a^2)(-a) + (-a^3)/3 =

= 2a^3 - 2(a^3)/3 = [4/3](a^3)

Area of the triangle = [1/2]base*height = [1/2](2a)(a)^2 = <span>a^3

ratio area of the triangle / area of the parabolic region = a^3 / {[4/3](a^3)} =

Limit of </span><span><span>a^3 / {[4/3](a^3)} </span>as a -> 0 = 1 /(4/3) = 4/3
</span>

3 0

3 years ago

In 2019 Lee began working as a nurse anesthetist with a taxable income of $142,545. The Medicare tax was 1.45% and the Social Se

9966 [12]

Answer:

free points

Step-by-step explanation:

thanksssss for the points

4 0

3 years ago

What is the diameter, radius, r2, and circumference?<br> 3 cm

Allushta [10]

Answer:

see below

Step-by-step explanation:

given:

radius of a circle = 3cm

find:

1. diameter

2. radius

3. r²

1) diameter = twice the radius = 2 r = 2 (3) = 6 cm

2) radius = 3 cm

3) r² = 3² = 9 cm

8 0

3 years ago

Pls help (15 pts)<br> whats the value of p

svetlana [45]

I think p will 2 is this 8th grade math ?

5 0

3 years ago