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WINSTONCH [101]
3 years ago
9

The following graph plots the batting averages of 29 players of a baseball team against the weekly time (in hours) they spend in

batting practice.
Which statement is a reasonable conclusion that may be drawn from the graph?

Practice and batting average have no direct relationship.
The best batter has an average of 0.33.
Batting practice increases batting averages.
The worst batter has an average of 0.

Mathematics
1 answer:
grin007 [14]3 years ago
7 0

Answer:

Batting practice increases batting averages.

Step-by-step explanation:

By looking at the graph, we clearly see a trend, almost a straight line (if you regroup the points a bit) going up.

So, it's clear that batting practice increases batting averages; they're directly related.  The more a player practices at bat, the better will be his average.

<u>Answers B and D </u>(best and worst batters) are not proved at all by the graph, since there's data higher than those two cases.

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A student simplifies (6b + 4r) – (2b + r) and says that the result is 4b + 5r. Explain the error and describe the correct steps
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Actually, its not 4b + 5r  its 4b +3r because look 

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step 2. 6b + 4r - 2b - r
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step 4. 4b + 3r  
So the answer is actually 4b + 3r 

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Pls help!!!!!!!;( i’ll give brainliest
olya-2409 [2.1K]

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Slav-nsk [51]

\huge \boxed{\mathfrak{Question} \downarrow}

Factorise the polynomials.

\large \boxed{\mathfrak{Answer \: with \: Explanation} \downarrow}

<h3><u>1. b² + 8b + 7</u></h3>

\sf \: b ^ { 2 } + 8 b + 7

Factor the expression by grouping. First, the expression needs to be rewritten as b²+pb+qb+7. To find p and q, set up a system to be solved.

\sf \: p+q=8 \\  \sf pq=1\times 7=7

As pq is positive, p and q have the same sign. As p+q is positive, p and q are both positive. The only such pair is the system solution.

\sf \: p=1  \\  \sf \: q=7

Rewrite \sf\:b^{2}+8b+7 \: as \: \left(b^{2}+b\right)+\left(7b+7\right).

\sf \: \left(b^{2}+b\right)+\left(7b+7\right)

Take out the common factors.

\sf \: b\left(b+1\right)+7\left(b+1\right)

Factor out common term b+1 by using distributive property.

\boxed{\boxed{ \bf \: \left(b+1\right)\left(b+7\right) }}

__________________

<h3><u>2. 4x² + 4x + 1</u></h3>

\sf \: 4 x ^ { 2 } + 4 x + 1

Factor the expression by grouping. First, the expression needs to be rewritten as 4x²+ax+bx+1. To find a and b, set up a system to be solved.

\sf \: a+b=4 \\  \sf ab=4\times 1=4

As ab is positive, a and b have the same sign. As a+b is positive, a and b are both positive. List all such integer pairs that give product 4.

\sf \: 1,4 \\  \sf 2,2

Calculate the sum for each pair.

\sf \: 1+4=5  \\  \sf \: 2+2=4

The solution is the pair that gives sum 4.

\sf \: a=2 \\  \sf b=2

Rewrite \sf4x^{2}+4x+1 as \left(4x^{2}+2x\right)+\left(2x+1\right).

\sf \: \left(4x^{2}+2x\right)+\left(2x+1\right)

Factor out 2x in 4x² + 2x.

\sf \: 2x\left(2x+1\right)+2x+1

Factor out common term 2x+1 by using distributive property.

\sf\left(2x+1\right)\left(2x+1\right)

Rewrite as a binomial square.

\boxed{ \boxed{\bf\left(2x+1\right)^{2}} }

__________________

<h3><u>3. 5n² + 10n + 20</u></h3>

\sf \: 5 n ^ { 2 } + 10 n + 20

Factor out 5. Polynomial n² + 2n+4 is not factored as it does not have any rational roots.

\boxed{\boxed{\bf5\left(n^{2}+2n+4\right)} }

<u>___________</u>_______

<h3><u>4. m³ - 729</u></h3>

\sf \: m ^ { 3 } - 729

Rewrite m³-729 as m³ - 9³. The difference of cubes can be factored using the rule: a^{3}-b^{3}=\left(a-b\right)\left(a^{2}+ab+b^{2}\right). Polynomial m²+9m+81 is not factored as it does not have any rational roots.

\boxed{ \boxed{ \bf \: \left(m-9\right)\left(m^{2}+9m+81\right) }}

__________________

<h3><u>5. x² - 81</u></h3>

\sf \: x ^ { 2 } - 81

Rewrite x²-81 as x² - 9². The difference of squares can be factored using the rule: a^{2}-b^{2}=\left(a-b\right)\left(a+b\right).

\boxed{ \boxed{ \bf\left(x-9\right)\left(x+9\right) }}

__________________

<h3><u>6. 15x² - 17x - 4</u></h3>

\sf \: 15 x ^ { 2 } - 17 x - 4

Factor the expression by grouping. First, the expression needs to be rewritten as 15x²+ax+bx-4. To find a and b, set up a system to be solved.

\sf \: a+b=-17 \\   \sf \: ab=15\left(-4\right)=-60

As ab is negative, a and b have the opposite signs. As a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -60.

\sf \: 1,-60  \\ \sf 2,-30  \\  \sf3,-20  \\  \sf4,-15 \\   \sf5,-12  \\ \sf 6,-10

Calculate the sum for each pair.

\sf \: 1-60=-59  \\ \sf 2-30=-28 \\   \sf3-20=-17 \\   \sf4-15=-11  \\  \sf5-12=-7  \\  \sf6-10=-4

The solution is the pair that gives sum -17.

\sf \: a=-20  \\  \sf \: b=3

Rewrite \sf15x^{2}-17x-4 as \sf \left(15x^{2}-20x\right)+\left(3x-4\right).

\sf\left(15x^{2}-20x\right)+\left(3x-4\right)

Factor out 5x in 15x²-20x.

\sf \: 5x\left(3x-4\right)+3x-4

Factor out common term 3x-4 by using distributive property.

\boxed{\boxed{ \bf\left(3x-4\right)\left(5x+1\right) }}

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