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3 years ago

I need help!!! due by 12

Mathematics

1 answer:

7nadin3 [17]3 years ago

4 0

Answer:

1. $11747.083

2. $29,253

3. $24222.04

4. $1542.23

Step-by-step explanation:

Message for explanation if you want.

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The Thomas Supply Company Inc. is a distributor of gas-powered generators. As with any business, the length of time customers ta

Dmitry [639]

Answer:

The answer is below

Step-by-step explanation:

When computing quartile and decile, the data must be arranged in ascending order.

Given the data points:

13 13 13 20 26 26 29 31 34 34 35 35 36 37 38 41 41 41 42 43 46 47 48 49 53 55 56 62 67 82

The numbers are arranged in ascending order. The total number of terms is 30.

$First \ quartile(Q_1)=\frac{1}{4}(n+1)^{th}\ term\\\\where\ n=number\ of\ terms. Hence:\\\\Q_1=\frac{1}{4}(30+1)=7.75^{th}\ term=average\ of\ 7th\ and\ 8th\ term=\frac{29+31}{2} =30\\\\Q_1=30\\\\Third \ quartile(Q_3)=\frac{3}{4}(n+1)^{th}\ term\\\\Q_3=\frac{3}{4}(30+1)=23.25^{th}\ term=average\ of\ 23rd\ and\ 24th\ term=\frac{48+49}{2} =48.5\\\\Q_3=48.5$

$Second \ decile(D_2)=\frac{2}{10}(n+1)^{th}\ term\\\\where\ n=number\ of\ terms. Hence:\\\\D_2=\frac{2}{10}(30+1)=6.2^{th}\ term=average\ of\ 6th\ and\ 7th\ term=\frac{26+29}{2} =27.5\\\\D_2=27.5\\\\\\Eight \ decile(D_8)=\frac{8}{10}(n+1)^{th}\ term=\frac{8}{10}(30+1)=24.8^{th}\ term\\=average\ of\ 24th\ and\ 25th\ term=\frac{49+53}{2} =51\\\\D_8=51$

6 0

3 years ago

P(x)= 3x^3-5x^2-14x-4

nexus9112 [7]

$\displaystyle\\
P(x)=3x^3-5x^2-14x-4\\\\
D_{-4}=\{-4;~-2;~\underline{\bf -1};~1;~2;~4\}\\\\
\text{We observe that } \frac{-1}{3} \text{ is a solution of the equation:}\\
3x^3-5x^2-14x-4=0\\\\
$

$\displaystyle\\
\text{Verification}\\\\
3x^3-5x^2-14x-4=\\\\
=3\times\Big(-\frac{1}{3}\Big)^3-5\times\Big(-\frac{1}{3}\Big)^2-14\times\Big(-\frac{1}{3}\Big)-4=\\\\
=-\frac{1}{9}-\frac{5}{9}+\frac{14}{3}-4=\\\\
=-\frac{6}{9}+\frac{14}{3}-4=\\\\
=-\frac{6}{9}+\frac{42}{9}- \frac{4\times 9}{9}=\\\\
 =-\frac{6}{9}+\frac{42}{9}- \frac{36}{9}= \frac{42-6-36}{9}=\frac{42-42}{9}=\frac{0}{9}=0\\\\
\Longrightarrow~~~P(x)~\vdots~\Big(x+ \frac{1}{3}\Big)\\\\
\Longrightarrow~~~P(x)~\vdots~(3x+1)$

$\displaystyle\\
3x^3-5x^2-14x-4=0\\
~~~~~-5x^2 = x^2 - 6x^2\\
~~~~~-14x =-2x-12x \\
3x^3+x^2 - 6x^2-2x-12x-4=0\\
x^2(3x+1)-2x(3x+1) -4(3x+1)=0\\
(3x+1)(x^2-2x -4)=0\\\\
\text{Solve: } x^2-2x -4=0\\\\
x_{12}= \frac{-b\pm \sqrt{b^2-4ac}}{2a}=\\\\=\frac{2\pm \sqrt{4+16}}{2}=\frac{2\pm \sqrt{20}}{2}=\frac{2\pm 2\sqrt{5}}{2}=1\pm\sqrt{5}\\\\
x_1 =1+\sqrt{5}\\
x_2 =1-\sqrt{5}\\
\Longrightarrow P(x)= 3x^3-5x^2-14x-4 =\boxed{(3x+1)(x-1-\sqrt{5})(x-1+\sqrt{5})}$

7 0

3 years ago

Solve for r.<br> –5 ≥ 5 (r+5)

aleksandr82 [10.1K]

Answer:

-6 ≥ r

Step-by-step explanation:

–5 ≥ 5 (r+5)

Divide each side by 5

–5/5 ≥ 5/5 (r+5)

–1 ≥ (r+5)

Subtract 5 from each side

-1-5 ≥ r+5-5

-6 ≥ r

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