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3 years ago

84 students can fit on 2 busses. How many students can fit on 9 busses

Mathematics

1 answer:

Julli [10]3 years ago

7 0

Answer:378 people

Step-by-step explanation:

84/2=42

42*9=378

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Use the properties of exponents to enter an equivalent expression with a single exponent.

Gelneren [198K]

$({10}^{2} ) {}^{4} = {10}^{2 \times 4} = {10}^{8}$

I hope this can help you ^_^

5 0

3 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,

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

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

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

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

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

7 0

3 years ago

−8 3/4÷2 1/6 please show me ur work

nalin [4]

Answer:

- 4 1/26

Step-by-step explanation:

<u>Solving in steps:</u>

- 8 3/4 ÷ 2 1/6 =
- 35/4 ÷ 13/6 =
- 35/4 × 6/13 =
- 35/2 × 3/13 =
- 105/26 =
- 4 1/26

6 0

3 years ago

Solve the formula d = rt for t. A. t = rd B. t = d – r C. t = D. t =

tekilochka [14]

$rt=d\qquad\text{divide both sides by}\ r\neq0\\\\\dfrac{rt}{r}=\dfrac{d}{r}\\\\\boxed{t=\dfrac{d}{r}}$

3 0

3 years ago

Read 2 more answers

A diver begins at 140 feet below sea level. She descends at a steady rate of 7 feet per minute for 4.5 minutes. Then, she ascend

Brut [27]

Answer:

Step-by-step explanation:

starting point: 140 feet below sea level.=-140

she then decends= 7(4.5)=31.5

-140-31.5=-171.5

finally she ascends 112.2 feet

-171.5+112.2=-59.3 feet or 59.3 feet below sea level

4 0

3 years ago