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2 years ago

13

a cable installer can make 42 installations in 6 hours how long will the installers been making a hundred and forty-seven instal

lations

2 answers:

Wewaii [24]2 years ago

5 0

This problem can solve by the rule of three.
42 installations-----------------------6 hours
147 installations---------------------- x

x=(147 installations * 6 hour) / 42 installations=21 hours.

Answer: 21 hours.

ipn [44]2 years ago

3 0

147/42=3.5*6=21
21 hours is your answer

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The distribution of heights of young women aged 18 to 24 is approximately normal with a mean of 64.5 inches and a standard devia

Lena [83]

The 68 - 95 - 99.7 rule, gives the basis to solve this question.

It says that for a normal distribution 95% of the results are between the mean minus 2 standard deviations and the mean plus 2 standard deviations.

Here:
mean = 64.5 inches,
standard deviaton = 2.5 inches

mean - 2 standard deviations = 64.5 inches - 5 inches = 59.5 inches

mean + 2 standard deviations = 64.5 inches + 5 inches = 69.5 inches

Then, the answer is that 95% of women range approximately between 59.5 inches and 69.5 inches.

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3 years ago

The length of a rectangle is 9 cm more then half the width. Find the length of the perimeter is 60cm

stira [4]

length = 1/2 w+9

perimeter= 60 =2(l+w)

substitute in for length

60 = 2(1/2 w +9 +w)

60 = 2 (3/2 w +9)

distribute

60 = 3w + 18

subtract 18 from each side

42 = 3w

divide by 3 on each side

14 = w

length = 1/2 w + 9

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7 + 9

16cm

7 0

3 years ago

In a G.P the difference between the 1st and 5th term is 150, and the difference between the

liubo4ka [24]

Answer:

Either $\displaystyle \frac{-1522}{\sqrt{41}}$ (approximately $-238$ ) or $\displaystyle \frac{1522}{\sqrt{41}}$ (approximately $238$ .)

Step-by-step explanation:

Let $a$ denote the first term of this geometric series, and let $r$ denote the common ratio of this geometric series.

The first five terms of this series would be:

$a$ ,
$a\cdot r$ ,
$a \cdot r^2$ ,
$a \cdot r^3$ ,
$a \cdot r^4$ .

First equation:

$a\, r^4 - a = 150$ .

Second equation:

$a\, r^3 - a\, r = 48$ .

Rewrite and simplify the first equation.

$\begin{aligned}& a\, r^4 - a \\ &= a\, \left(r^4 - 1\right)\\ &= a\, \left(r^2 - 1\right) \, \left(r^2 + 1\right) \end{aligned}$ .

Therefore, the first equation becomes:

$a\, \left(r^2 - 1\right) \, \left(r^2 + 1\right) = 150$ ..

Similarly, rewrite and simplify the second equation:

$\begin{aligned}&a\, r^3 - a\, r\\ &= a\, \left( r^3 - r\right) \\ &= a\, r\, \left(r^2 - 1\right) \end{aligned}$ .

Therefore, the second equation becomes:

$a\, r\, \left(r^2 - 1\right) = 48$ .

Take the quotient between these two equations:

$\begin{aligned}\frac{a\, \left(r^2 - 1\right) \, \left(r^2 + 1\right)}{a\cdot r\, \left(r^2 - 1\right)} = \frac{150}{48}\end{aligned}$ .

Simplify and solve for $r$ :

$\displaystyle \frac{r^2+ 1}{r} = \frac{25}{8}$ .

$8\, r^2 - 25\, r + 8 = 0$ .

Either $\displaystyle r = \frac{25 - 3\, \sqrt{41}}{16}$ or $\displaystyle r = \frac{25 + 3\, \sqrt{41}}{16}$ .

Assume that $\displaystyle r = \frac{25 - 3\, \sqrt{41}}{16}$ . Substitute back to either of the two original equations to show that $\displaystyle a = -\frac{497\, \sqrt{41}}{41} - 75$ .

Calculate the sum of the first five terms:

$\begin{aligned} &a + a\cdot r + a\cdot r^2 + a\cdot r^3 + a \cdot r^4\\ &= -\frac{1522\sqrt{41}}{41} \approx -238\end{aligned}$ .

Similarly, assume that $\displaystyle r = \frac{25 + 3\, \sqrt{41}}{16}$ . Substitute back to either of the two original equations to show that $\displaystyle a = \frac{497\, \sqrt{41}}{41} - 75$ .

Calculate the sum of the first five terms:

$\begin{aligned} &a + a\cdot r + a\cdot r^2 + a\cdot r^3 + a \cdot r^4\\ &= \frac{1522\sqrt{41}}{41} \approx 238\end{aligned}$ .

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2 years ago

Polynomial Warm Up

Kazeer [188]

Answer:

Step-by-step explanation:

False you must multiply the coefficient (numbers) and add the exponents.

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3 years ago

Yuki888 [10]

R'S' is equal in length to RS.

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2 years ago

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