Ask question

Ask question

All categories

kondor19780726 [428]

2 years ago

8

Ann took a taxi home from the airport. The taxi fare was $2.10 per mile, and she gave the driver a tip of $5. Ann paid a total o

f $49.10.
Write an equation to determine the distance in miles (x) between the airport and Ann's home
Find the distance between the airport and Ann's home.

1 answer:

adell [148]2 years ago

3 0

Answer:

$2.10x + 5= $49.10

x=21

Step-by-step explanation:

the x is used for a variable that changes and in this case the mph changes which lead x being the amount of miles that they had drove which at the end ends up multi. with the $2.10 . as for the 5 it was an initial as in it doesn't change.

for the second part

you subtract 5 from each side so you have the total of 2.10x= 44.10 then u divide 2.10 from each side and that will give you x = 21

You might be interested in

Solve: 5/12 × 8/15=

Cerrena [4.2K]

Answer:0.5

Step-by-step explanation: If you flip the numbers and the continue to multiply them across and then divide them after you should get 0.5

8 0

3 years ago

Plz help fast!!!!!!

Leokris [45]

Answer:

C

Step-by-step explanation:

7 0

2 years ago

Read 2 more answers

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

7 1/2 - 6 3/10 find the differnce

tigry1 [53]

Answer: they are different numbers

Step-by-step explanation:

7 0

2 years ago

HELP ME PLEASE !!!!

lyudmila [28]

For this one, an easy way to solve it is to break it down into parts. First lets convert 5 feet to inches.

Multiply the number of feet by how many inches are in one foot.

5 times 12= 60 in

Now just add the like terms

60 in + 2 in= 62 inches.

Now we see that Sarah is 62 inches tall, now we can convert to centimeters the same way.

62 times 2.54= 157.48

So that makes C your answer, Sarah is 157.48 cm tall.

5 0

3 years ago