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1 year ago

Find the mean, median and mode of the weights of the people shown. 105kg 53kg 76kg 91kg 120kg 61kg 55kg 98kg 61kg

2 answers:

5 0

First, you need to put them in order
53kg, 55kg, 61kg, 61kg, 76kg, 91kg, 98kg, 105kg, 120kg

For mean, you add them all up and divide by the amount of numbers (9)
720/9 = 80

For median, you find the middle number (76kg)

For mode, you find the number that appears the most (61kg)

Mean: 80
Median: 76
Mode: 61

svetoff [14.1K]1 year ago

5 0

Mean = 80kg

Add all the numbers together and devide by how many

Median = 76kg

Find the middle number

Mode =61kg

The most frequent number, the number that appears the most.

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Nevin made tables of values to approximate the solution to a system of equations. First he found that the x-value of the solutio

prisoha [69]

Answer:

The correct option is B.

Step-by-step explanation:

The given equation are

$y=4x-3$

$y=-5x+9$

On solving both the equation, we get

$4x-3=-5x+9$

$4x+5x=9+3$

$9x=12$

$9x=\frac{12}{9}=\frac{4}{3}$

Put this value in the given equation.

$y=4(\frac{4}{3})-3$

$y=\frac{16}{3}-3$

$y=\frac{16-9}{3}$

$y=\frac{7}{3}$

The solution of the given system of equation is

$(\frac{4}{3},\frac{7}{3})=(1.333,2.333)\approx (1.3,2.3)$

The best approximation of the exact solution is (1.3,2.3). Therefore the correct option is B.

7 0

2 years ago

Read 2 more answers

A pipe that is 120 cm long resonates to produce sound of wavelengths 480 cm, 160 cm, and 96 cm but does not resonate at any wave

erma4kov [3.2K]

Answer:

A Pipe that is 120 cm long resonates to produce sound of wavelengths 480 cm, 160 cm and 96 cm but does not resonate at any wavelengths longer than these. This pipe is:

A. closed at both ends

B. open at one end and closed at one end

C. open at both ends.

D. we cannot tell because we do not know the frequency of the sound.

The right choice is:

B. open at one end and closed at one end .

Step-by-step explanation:

Given:

Length of the pipe, $L$ = 120 cm

Its wavelength $\lambda_1$ = 480 cm

$\lambda_2$ = 160 cm and $\lambda_3$ = 96 cm

We have to find whether the pipe is open,closed or open-closed or none.

Note:

The fundamental wavelength of a pipe which is open at both ends is 2L.
The fundamental wavelength of a pipe which is closed at one end and open at another end is 4L.

So,

The fundamental wavelength:

⇒ $4L=4(120)=480\ cm$

It seems that the pipe is open at one end and closed at one end.

Now lets check with the subsequent wavelengths.

For one side open and one side closed pipe:

An odd-integer number of quarter wavelength have to fit into the tube of length L.

⇒ $\lambda_2=\frac{4L}{3}$ ⇒ $\lambda_3=\frac{4L}{5}$

⇒ $\lambda_2=\frac{4(120)}{3}$ ⇒ $\lambda_3=\frac{4(120)}{5}$

⇒ $\lambda_2=\frac{480}{3}$ ⇒ $\lambda_3=\frac{480}{5}$

⇒ $\lambda_2=160\ cm$ ⇒ $\lambda_3=96\ cm$

So the pipe is open at one end and closed at one end .

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1 year ago

Brad owns 3 surfboards and 2 wet suits. If he takes one surfboard and one wet suit to the beach, how many different combinations

Reptile [31]

Answer:

Step-by-step explanation:

He only has 1 wetsuit and 1 surfboard.

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1 year ago

Jennifer has a bag of chips that contains 2 red chips and 1 black chip. If she also has a fair die, what is the probability that

fredd [130]

<span> the probability that she rolls an odd number AND and pulls a red chip

so it is = Prob(odd no) * Prob(red chip)

Prob(odd no) for a fair die = 1/2

Prob(red chip) = red chip / total chip = 2/(2+1) = 2/3

so the ans is 1/2 * 2/3 = 1/3

</span>

7 0

1 year ago

Read 2 more answers

$1.12 for 8.2 ounces? Unit Rate

Reil [10]

That is around 13 or 14 cents. The specific answer is 13.65853659 cents, but you can either round up to 14 or round down to 13 (rounding up to 14 is the correct way to do it.)

First you divide 1.12 by 8.2 to get the unit rate because It takes $1.12 to buy 8.2 ounces. 1.12/8.2=13.65853659

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1 year ago