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3 years ago

What is 800 metres in centimetres

Mathematics

2 answers:

xxTIMURxx [149]3 years ago

7 0

There are 100 centimeters in on meter so multiply 800*100=80000. so you know that there are 80,000 cm in 800 meters.

KIM [24]3 years ago

3 0

Remember that
1m=100cm
8m=800cm

800m=80,000cm

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In the past, the average age of employees of a large corporation has been 40 years. Recently, the company has been hiring older

earnstyle [38]

Answer:

Step-by-step explanation:

In the past, mean of age of employees

i.e. $\mu = 40$

Recently sample was taken

n = sample size = 60

Mean of sample = 45

Std dev of sample s = 16

$H_0: \bar x = 40\\H_a: \bar x >40$

(Right tailed test)

Since only population std deviation is known we can use t test only

Std error = $\frac{s}{\sqrt{n} } \\=\frac{16}{\sqrt{61} } \\=2.049$

Mean difference = 45-40 =5

Test statistic t= $\frac{5}{2.049} \\=2.441$

df = 60

p value =0.008739

Since p < 0.05 we reject null hypothesis

The mean age has increased.

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3 years ago

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Vinvika [58]

Answer:

Step-by-step explanation:

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2 years ago

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AysviL [449]

Answer: <less than >greater than ≥ greater than or equal to ≤ less than or equal to

Step-by-step explanation:

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3 years ago

True or false?

Nata [24]

True...

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2 years ago

Help i dont understand how to complete this

Luda [366]

Answer: Josh = 19.8 hours, Danny = 10.8 hours

Step-by-step explanation:

$Josh: \dfrac{1}{x+9}\\\\\\Danny: \dfrac{1}{x}\\\\\\Together: \dfrac{1}{7}\\\\\\Josh\quad + \quad Danny\quad =\quad Together\\\dfrac{1}{x+9}\quad +\qquad \dfrac{1}{x}\qquad = \qquad \dfrac{1}{7}\\\\\\\underline{\text{Clear the denominator by multiplying by the LCM: (x+9)(x)(7)}}\\\\7(x) + 7(x+9)=x(x+9)\\7x + 7x + 63 = x^2+9x\\14x+63=x^2+9x\\0=x^2-5x-63\\\\\\\underline{\text{Use the quadratic formula to solve for x: }}x=\dfrac{-b\pm \sqrt{b^2-4ac}}{2a}$

$x=\dfrac{-(-5)\pm \sqrt{(-5)^2-4(1)(-63)}}{2(1)}\\\\\\.\quad =\dfrac{5\pm\sqrt{25+252}}{2}\\\\\\.\quad =\dfrac{5\pm\sqrt{277}}{2}\\\\\\.\quad =\dfrac{5\pm 16.6}{2}\\\\\\x =\dfrac{5+16.6}{2}\qquad x=\dfrac{5-16.6}{2}\\\\\\x=\dfrac{21.6}{2}\qquad \qquad x=\dfrac{-11.6}{2}\\\\\\x=10.8 \qquad \qquad x=-5.8$

Since time cannot be negative, x=-5.8 is an extraneous solution (not valid) so x = 10.8

Josh: x + 9 --> 10.8 + 9 = 19.8

Danny: x --> 10.8

****************************************************************************************

2a) k = 9.45 & 0.55

2b) x = 3/2

2c) x = 2/3 (x = -1 is an extraneous solution so is not valid)

2d) No Solution (x = 1 is an extraneous solution so is not valid)

Here is the work for 2a. Follow this format for b, c, & d

$\dfrac{3}{k^2-8x+12}=\dfrac{k}{k-2}-\dfrac{4}{k-6}\\\\\text{Since the denominator cannot equal zero, then } k \neq2\ and\ k\neq6\\\text{If either of the solutions are 2 or 6, then that solution needs to be crossed out}\\\\\underline{\text{Clear the denominator by multiplying by the LCM: (k-2)(k-6)}}\\\\3 = k(k-6)-4(k-2)\\3=k^2-6k-4k+8\\0=k^2-10k+5\\\\\\\underline{\text{Use the quadratic formula to solve for k}}$

$x=\dfrac{-(-10)\pm \sqrt{(-10)^2-4(1)(5)}}{2(1)}\\\\\\.\quad =\dfrac{10\pm\sqrt{100-20}}{2}\\\\\\.\quad =\dfrac{10\pm\sqrt{80}}{2}\\\\\\.\quad =\dfrac{10\pm8.9}{2}\\\\\\x=\dfrac{10+8.9}{2}\qquad x=\dfrac{10-8.9}{2}\\\\\\x=\dfrac{18.9}{2}\qquad x=\dfrac{1.1}{2}\\\\\\x=9.45\qquad x=0.55\\\\\\\text{Both answers are valid!}$

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3 years ago