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3 years ago

Multiply the following numbers. Reduce the answer to lowest terms. 3 1/5 x 2 1/7

Mathematics

2 answers:

PSYCHO15rus [73]3 years ago

8 0

The answer is A. 6 6/7

Mashcka [7]3 years ago

6 0

Answer:

Step-by-step explanation:

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the area of a trapezoid is 36 square inches. The height is 4 inches and one bases twice the length of the other base. What are t

vlada-n [284]

The formula for this is $A= \frac{1}{2} ( b_{1} + b_{2} )h$ . Filling in accordingly, $36= \frac{1}{2}(x+2x)(4)$ . This simplifies to $36= \frac{1}{2}(3x)(4)$ and 36=6x. Therefore, x = 6 so the bases are 6 and 12.

8 0

2 years ago

Read 2 more answers

Anyone know what the answer is?

babymother [125]

A. 12

B. 34

C. (x+2)+(x+2)+(5+5)

3 0

2 years ago

PLEASE HELP MEEEEEE PLEASEEEEEEEE!!!!!! IM TIMED

crimeas [40]

Set up an equation to solve.

$8.50x (since it is the rate of change) + $12 (flat fee) = total money earned (or y). In this case, we have the total, amount, and are looking for x. Plug $139.50 into the equation.

$8.50x + $12 = $139.50

Now, solve for x.

$8.50x + $12-12 = $139.50-12 > $8.50x = $127.5

$8.50/8.50x = $127.5/8.50 > x = 15

So, the correct answer is D. Olivia babysat for 15 hours.

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3 years ago

Solve the inequality 15t > 180

bagirrra123 [75]

Answer:

t>12

Step-by-step explanation:

t>180/15

then divide 180 by 15

and you gey t>12

4 0

2 years ago

Helppp me plsssssssss

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,

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

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

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

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

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

7 0

3 years ago