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2 years ago

June 14th Jim Thompson Homework

Mathematics

2 answers:

Oduvanchick [21]2 years ago

8 0

Answers:

8) Choices E, F and G
9) r = -0.95
10) 0.9 (choice C)
11) Choice C. y1 = mx1+b
12) Choice B. Slope
13) Choice C. y intercept

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Explanations:

8) Positive r values always go with positive regression slopes. The r value will likely not equal the slope, but they share the same sign.

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9) The data set isn't given so I'm not sure. But it seems like your teacher just wants to know a hypothetical situation in which r value will produce the best fit. That would be r = -0.95 since we're the furthest from zero. The closer you are to r = 1 or r = -1, the better the fit.

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10) The blue regression line, aka line of best fit, is fairly close to all of the data points. Also, the blue line has a positive slope. This means r is fairly close to 1. So that's why we go with r = 0.9

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11) The general equation is y = mx+b. We simply replace x,y with x1,y1.

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12) When considering y = mx+b, the m is the slope. I'm not entirely sure where the m comes from, but I've read some sources that say 'm' somehow represents 'mountain' which involves slopes. This could have simply been a memorization tool and not the actual reason why m was chosen to be the slope.

As an example, y = 2x+5 has slope 2.

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13) The b in y = mx+b is the y intercept. It's the initial value.

It always occurs when x = 0 since

y = mx+b

y = m*0+b

y = b

Meaning the point (0,b) is the location of the y intercept.

For example, y = 2x+5 has y intercept 5 located at the point (0,5).

pogonyaev2 years ago

3 0

Seriously, how do those two images put on brainly? :/

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Find the sum of the positive integers less than 200 which are not multiples of 4 and 7

taurus [48]

Answer:

$12942$ is the sum of positive integers between $1$ (inclusive) and $199$ (inclusive) that are not multiples of $4$ and not multiples $7$ .

Step-by-step explanation:

For an arithmetic series with:

$a_1$ as the first term,
$a_n$ as the last term, and
$d$ as the common difference,

there would be $\displaystyle \left(\frac{a_n - a_1}{d} + 1\right)$ terms, where as the sum would be $\displaystyle \frac{1}{2}\, \displaystyle \underbrace{\left(\frac{a_n - a_1}{d} + 1\right)}_\text{number of terms}\, (a_1 + a_n)$ .

Positive integers between $1$ (inclusive) and $199$ (inclusive) include:

$1,\, 2,\, \dots,\, 199$ .

The common difference of this arithmetic series is $1$ . There would be $(199 - 1) + 1 = 199$ terms. The sum of these integers would thus be:

$\begin{aligned}\frac{1}{2}\times ((199 - 1) + 1) \times (1 + 199) = 19900 \end{aligned}$ .

Similarly, positive integers between $1$ (inclusive) and $199$ (inclusive) that are multiples of $4$ include:

$4,\, 8,\, \dots,\, 196$ .

The common difference of this arithmetic series is $4$ . There would be $(196 - 4) / 4 + 1 = 49$ terms. The sum of these integers would thus be:

$\begin{aligned}\frac{1}{2}\times 49 \times (4 + 196) = 4900 \end{aligned}$

Positive integers between $1$ (inclusive) and $199$ (inclusive) that are multiples of $7$ include:

$7,\, 14,\, \dots,\, 196$ .

The common difference of this arithmetic series is $7$ . There would be $(196 - 7) / 7 + 1 = 28$ terms. The sum of these integers would thus be:

$\begin{aligned}\frac{1}{2}\times 28 \times (7 + 196) = 2842 \end{aligned}$

Positive integers between $1$ (inclusive) and $199$ (inclusive) that are multiples of $28$ (integers that are both multiples of $4$ and multiples of $7$ ) include:

$28,\, 56,\, \dots,\, 196$ .

The common difference of this arithmetic series is $28$ . There would be $(196 - 28) / 28 + 1 = 7$ terms. The sum of these integers would thus be:

$\begin{aligned}\frac{1}{2}\times 7 \times (28 + 196) = 784 \end{aligned}$ .

The requested sum will be equal to:

the sum of all integers from $1$ to $199$ ,
minus the sum of all integer multiples of $4$ between $1\!$ and $199\!$ , and the sum integer multiples of $7$ between $1$ and $199$ ,
plus the sum of all integer multiples of $28$ between $1$ and $199$ - these numbers were subtracted twice in the previous step and should be added back to the sum once.