Find the mean median mode and range of this date 49,49,54,55,52,49,55, if necessary round to the nearest tenth

1 answer:

6 0

The numbers from least to greatest is (49, 49, 49, 52, 54, 55, 55) The mean is 51.8 (52 rounded), 52 is median, mode is 49, and the range is 6. :)

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Can a 14 inch umbrella fit inside a suitcase that has dimensions of 9" x 12"? Explain.

inessss [21]

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5 0

2 years ago

Read 2 more answers

Solve the system of equations by substitution. 3/8 x + 1/3 y =17/24 and <br> x + 7y = 8

goblinko [34]

$\left \{ {{ \frac{3}{8}x+ \frac{1}{3}y = \frac{17}{14} } \atop {x+7y=8}} \right.$

To solve this system by substitution, first isolate x in the second equation.

$x+7y=8$
$x+7y-7y=8-7y$
$x=8-7y$

Now, plug this expression ( $8-7y$ ) for x in the top equation to solve for y.

$\frac{3}{8} (8-7y)+ \frac{1}{3} y= \frac{17}{14}$
$3- \frac{21}{8} y+ \frac{1}{3} y= \frac{17}{24}$
$72-63y+8y=17$
$72-55y=17$
$-55y=17-72$
$-55y=-55$
$y=1$

Now that you have y, plug it into the second equation and solve for x.

$x+7y=8$
$x+7(1)=8$
$x+7=8$
$x=1$

Last step is to plug your x- and y-values in to both equations to check your work.

$\frac{3}{8} (1)+ \frac{1}{3} y= \frac{17}{24}$

$\frac{3}{8} * \frac{3}{3} = \frac{9}{24} ; \frac{1}{3} * \frac{8}{8} = \frac{8}{24}$

$\frac{9}{24} + \frac{8}{24} = \frac{17}{24}$ <--True

$1+7(1)=8$
$1+7=8$ <--True

Answer:
$x=1 \\ y=1$

5 0

3 years ago

Sam use 6<img src="https://tex.z-dn.net/?f=%5Cfrac3%7B4%7D%7B%7D" id="TexFormula1" title="\frac3{4}{}" alt="\frac3{4}{}" align="

PtichkaEL [24]

$\bf \begin{array}{ccll} \stackrel{strawberry}{lbs}&\stackrel{jam}{cups}\\ \cline{1-2} \frac{3}{4}&\frac{1}{3}\\\\ x&1 \end{array}\implies \cfrac{~~\frac{3}{4}~~}{x}=\cfrac{~~\frac{1}{3}~~}{1}\implies \cfrac{~~\frac{3}{4}~~}{\frac{x}{1}}=\cfrac{1}{3}\implies \cfrac{3}{4}\cdot \cfrac{1}{x}=\cfrac{1}{3} \\\\\\ \cfrac{3}{4x}=\cfrac{1}{3}\implies 9=4x\implies \cfrac{9}{4}=x\implies 2\frac{1}{4}=x$