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2 years ago

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Match the vocabulary word with the correct definition. 1. circle circles that have equal radii 2. radius a segment joining the c

enter of a circle to a point of the circle 3. diameter the set of all points in a plane that are at a given distance from a given point in the plane 4. congruent circles two or more circles that lie in the same plane an have the same center 5. concentric circles a chord of a circle that contains the center of the circle

2 answers:

Rudik [331]2 years ago

8 0

What are your answer choices ?

Oduvanchick [21]2 years ago

4 0

Answer:

1 - Circle 2 - Concentric 3 - Congruent 4 - Diameter 5 - Radius

Step-by-step explanation:

1. the set of all points in a plane that are at a given distance from a given point in the plane

Circle

2. two or more circles that lie in the same plane and have the same center

Concentric Circles

3. circles that have equal radii

Congruent Circles

4. a chord of a circle that contains the center of the circle

Diameter

5. a segment joining the center of a circle to a point of the circle

Radius

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A firm buys 12 file cabinets at $140 each, with payment in full due in 90 days. How much must the firm deposit now to have enoug

anyanavicka [17]

Answer:

firm deposit is $1667.50

Step-by-step explanation:

given data

buys 12 file = $140

time = 90 days

interest = 3%

solution

we get here return on money over 90 day period is

we know Return on 360 days = 3%

return on money over 90 day = 3% ÷ 4 = 0.75%

we consider here required amount = x

so

$140 × 12 = x (1+0.0075)

x = $\frac{1680}{1+0.0075}$

x = 1667.50

so firm deposit is $1667.50

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2 years ago

Jack usually mows his lawn in 6 hours. marilyn can mow the same yard in 4 hours. How much time would it take for them to mow the

Anuta_ua [19.1K]

Answer:

2.4 hours or 2 hours and 24 minutes

Step-by-step explanation:

Jack mows 1/6 of the lawn per hour while Marilyn mows 1/4 of the lawn together. Adding up those rates yields in their combined mowing rate:

$R = \frac{1}{6}+ \frac{1}{4}\\R=\frac{10}{24}$

The time required for them to mow the entire lawn is:

$t = \frac{1}{R}\\t = \frac{1*24}{10}\\ t=2.4\ hours$

It would take them both 2.4 hours to mow the lawn together.

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2 years ago

Helppp me plsssssssss<br><br>

Oliga [24]

Answer:

The class 35 - 40 has maximum frequency. So, it is the modal class.

From the given data,

$\sf \:\:\:\:\:\:\:\:\:\:x_{k}=35$
$\sf \:\:\:\:\:\:\:\:\:\:f_{k}=50$
$\sf \:\:\:\:\:\:\:\:\:\:f_{k-1}=34$
$\sf \:\:\:\:\:\:\:\:\:\:f_{k+1}=42$
$\sf \:\:\:\:\:\:\:\:\:\:h=5$

${\bf \:\: {By\:using\:the\: formula}} \\ \\$

$\:\dag\:{\small{\underline{\boxed{\sf {Mode,\:M_{o} =\sf\red{x_k + {\bigg(h \times \: \dfrac{ ( f_k - f_{k-1})}{ (2f_k - f_{k - 1} - f_{k +1})}\bigg)}}}}}}} \\ \\$

$\sf \:\:\:\:\:\:\:\:\:= 35+ {\bigg(5 \times \dfrac{(50 - 34)}{ ( 2 \times 50 - 34 - 42)}\bigg)} \\ \\$

$\sf \:\:\:\:\:\:\:\:\:\:\:\:\:\:\:= 35 +{\bigg(5 \times \dfrac{16}{24}\bigg)} \\ \\$

$\sf \:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:= {\bigg(35+\dfrac{10}{3}\bigg)} \\ \\$

$\sf \:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:(35 + 3.33) =.38.33 \\ \\$

$\:\:\sf {Hence,}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\ \large{\underline{\mathcal{\gray{ mode\:=\:38.33}}}} \\ \\$

${\large{\frak{\pmb{\underline{Additional\: information }}}}}$

MODE

Most precisely, mode is that value of the variable at which the concentration of the data is maximum.

MODAL CLASS

In a frequency distribution the class having maximum frequency is called the modal class.

${\bf{\underline{Formula\:for\: calculating\:mode:}}} \\$

${\underline{\boxed{\sf {Mode,\:M_{o} =\sf\red{x_k + {\bigg(h \times \: \dfrac{ ( f_k - f_{k-1})}{ (2f_k - f_{k - 1} - f_{k +1})}\bigg)}}}}}} \\ \\$

Where,

$\sf \small\pink{ \bigstar} \: x_{k}= lower\:limit\:of\:the\:modal\:class\:interval.$

$\small \blue{ \bigstar}$ $\sf \: f_{k}=frequency\:of\:the\:modal\:class$

$\sf \small\orange{ \bigstar}\: f_{k-1}=frequency\:of\:the\:class\: preceding\:the\;modal\:class$

$\sf \small\green{ \bigstar}\: f_{k+1}=frequency\:of\:the\:class\: succeeding\:the\;modal\:class$

$\small \purple{ \bigstar}$ $\sf \: h= width \:of\:the\:class\:interval$

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