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Tcecarenko [31]

3 years ago

13

Baichung sfather is 2u years younger than Baichung s grandfather and 29 years older than Baichung.The sum of ages of as the thre

e is 135 years.What is the age of each of them?

2 answers:

juin [17]3 years ago

6 0

Here is the answer
hope this helps !!
mark me as brainliest ☺☺

Karo-lina-s [1.5K]3 years ago

6 0

Baichung sfather 54, Baichung s grandfather 56, Baichung 25.

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Help me solve this image please!!!

tresset_1 [31]

Based on the information, Christian would have $5525.5 of an annuity.

<h3>How to calculate the annuity?</h3>

According to the given information, the number of coffees per week is 3 then, per month is 3x4 = 12

Each coffee is $4.5. Then monthly expenditure for coffees is 12 x 4.5 = $54

Rate of interest r = 1.6% = 1.6/100 = 0.016 and for monthly compounding r = 0.016/12 = 0.00133

n = number of payments = 8 x 12 = 96

We can use the formula for finding the future value as below

FV = C x [ ( 1 + r )n-1 ] / ( r )

FV = 54 x [ ( 1 + 0.00133 )96 – 1 ] / (0.00133)

= 54 x [ (1.13609 - 1)] / (0.00133)

= 54 x 0.13609 / (0.00133)

= 54 x 102.3233

= 5525.5

Therefore Christian would have $5525.5 of the annuity.

Learn more about annuity on:

brainly.com/question/5303391

#SPJ1

5 0

2 years ago

14 - За = -а<br> What would be the answer?

LiRa [457]

Answer:

7

Step-by-step explanation:

collect like terms

14=-a+3a

2a=14

dividing both sides

a=7

5 0

3 years ago

Read 2 more answers

Tom wants to order tickets online so that he and three of his friends can go together to a water park. The cost of the tickets i

laiz [17]

16n + 2.50 = $18.50n

Will this do?

4 0

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,

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

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

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

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

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

7 0

3 years ago

Order least to greatest 1/2, 0.55, 5/7

Nataliya [291]

1/2<0.55<5/7
This is the order from least to greatest

8 0

3 years ago

Read 2 more answers