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notka56 [123]
3 years ago
6

Can you help me Solve this and show the stepd

Mathematics
1 answer:
NikAS [45]3 years ago
6 0
130-51 is 79 this is it D
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Which r-value represents the strongest correlation?<br> –0.83<br> –0.67<br> 0.48<br> 0.79
maks197457 [2]

Answer:

–0.83

Step-by-step explanation:

An r-value, or correlation coefficient, tells us the strength of the correlation in a linear regression.  This number ranges from -1 to 1; -1 is a perfect linear fit for a decreasing set of data, while 1 is a perfect linear fit for an increasing set of data.

The closer the r-value is to either -1 or 1, the stronger the correlation is.

The two negative numbers we have are -0.83 and -0.67.  The first one, -0.83, is 0.17 away from -1.  -0.67, on the other hand, is 0.33 away from -1.  The two positive numbers we have are 0.48 and 0.79.  The first one, 0.48, is 0.52 away from 1.  The second one, 0.79, is 0.21 away from 1.  The one that is closest to the perfect fit is -0.83, since it is only 0.17 away from a perfect fit.

8 0
2 years ago
Read 2 more answers
If f ( x ) = 2 x - 5, then f (4) is _____.<br><br> 17<br> 5<br> 4<br> 3<br><br> Thx
Slav-nsk [51]

Answer:

f(4)= 3

Step-by-step explanation:

f ( x ) = 2 x - 5

then f(4)= 2(4)-5

f(4)= 8-5

f(4)= 3

4 0
2 years ago
Read 2 more answers
<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
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