All categories

3 years ago

The fourth question please.

Mathematics

1 answer:

garik1379 [7]3 years ago

8 0

Going from the left to the right. 9, 4, 13, 15, 12, 6, 11. The total cost is the sum of these, or $70,000.

You might be interested in

HELP I HAVE TEN MINUTES!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

Flura [38]

No

Explanation

Do it on your own

8 0

2 years ago

Makaila read 44 pages of her book yesterday and 2/3 of the book today. If she read a total of 200 pages in the past two days how

pogonyaev

Answer:

Total number of pages in the book is 234.

Step-by-step explanation:

Let us assume that the total number of pages in the book = x

Now, it is given that Makaila read 44 pages yesterday.

Also, Makaila read two-third of the book today. As we have assumed that, there are a total of x number of pages in the book, so we can say that Makaila read two-third of x number of pages today.

Number of pages read by Makaila today = $\frac{2}{3}\;of\;x=\frac{2}{3}\times\;x=\frac{2x}{3}$

Now, it is also given that, Makaila read a total of 200 pages in the past two days.

∴ according to question,

Pages read yesterday + Pages read today = 200

⇒ $44+\frac{22x}{3}=200$

⇒ $\frac{44\times3+2x}{3}=200$

⇒ $\frac{132+2x}{3}=200$

⇒ $132+2x=200\times3\;\;\;\;\;\;\;\;\;[On\;cross-multiplying]$

⇒ $132+2x=600$

⇒ $2x=600-132$

⇒ $2x=468$

⇒ $x=\frac{468}{2}=234$

∴ Total number of pages in the book = x = 234

6 0

3 years ago

Help please i’ll give extra points

yKpoI14uk [10]

Answer:

3 units right and 1 unit up.

7 0

2 years ago

Read 2 more answers

Describe the following set as either finite or infinite.

serg [7]

Answer:

infinite

Step-by-step explanation:

6 0

3 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