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ArithmeticQuestion and Answers: Page 6

Question Number 202251 Answers: 1 Comments: 4

(1/(1×2×3)) + (1/(2×3×4)) + (1/(3×4×5)) + .............. + (1/(n(n+1)(n+2))) = ?

$\frac{1}{1 \times 2 \times 3} + \frac{1}{2 \times 3 \times 4} + \frac{1}{3 \times 4 \times 5} + . . . . . . . . . . . . . . + \frac{1}{n (n + 1) (n + 2)} = ?$

Question Number 202183 Answers: 1 Comments: 1

what is (√2) over 2

$w h a t i s \sqrt{2} o v e r 2$

Question Number 202154 Answers: 3 Comments: 1

⇃

$⇃ \underset{―}{}$

Question Number 202123 Answers: 2 Comments: 0

(1/(1×3)) + (1/(3×5)) + (1/(5×7)) + ...............∞ = ?

$\frac{1}{1 \times 3} + \frac{1}{3 \times 5} + \frac{1}{5 \times 7} + . . . . . . . . . . . . . . . \infty = ?$

Question Number 202086 Answers: 1 Comments: 0

There are two possible routes from Zindhi to Katifa. One route is through Zindhi/Chadler expressway which is 100km and the other is through Adfeti and Ngonu covering a distance of 80km. A motorist going through the expressway can travel 10km/h faster than the one going through Adfeti and Ngonu, and arrive Katifa 5 minutes earlier as well. What is the time spent on the journey to Katifa by the motorist travelling through the expressway.

$T h e r e a r e t w o p o s s i b l e r o u t e s f r o m$ $Z i n d h i t o K a t i f a . O n e r o u t e i s t h r o u g h$ $Z i n d h i / C h a d l e r e x p r e s s w a y w h i c h i s$ $100 k m a n d t h e o t h e r i s t h r o u g h A d f e t i a n d$ $N g o n u c o v e r i n g a d i s t a n c e o f 80 k m . A$ $m o t o r i s t g o i n g t h r o u g h t h e e x p r e s s w a y$ $c a n t r a v e l 10 k m / h f a s t e r t h a n t h e o n e$ $g o i n g t h r o u g h A d f e t i a n d N g o n u, a n d$ $a r r i v e K a t i f a 5 m i n u t e s e a r l i e r a s w e l l .$ $W h a t i s t h e t i m e s p e n t o n t h e j o u r n e y$ $t o K a t i f a b y t h e m o t o r i s t t r a v e l l i n g$ $t h r o u g h t h e e x p r e s s w a y .$

Question Number 202079 Answers: 1 Comments: 0

A man travelled from town A to B, a distance of 360km. He left A one hour later than he had planned so he decided to drive at 5km/h faster than his normal speed, in order to reach B on schedule. If he arrived B at exactly the scheduled time, find the normal speed.

$A m a n t r a v e l l e d f r o m t o w n A t o B, a$ $d i s t a n c e o f 360 k m . H e l e f t A o n e h o u r$ $l a t e r t h a n h e h a d p l a n n e d s o h e d e c i d e d$ $t o d r i v e a t 5 k m / h f a s t e r t h a n h i s$ $n o r m a l s p e e d, i n o r d e r t o r e a c h B o n$ $s c h e d u l e . I f h e a r r i v e d B a t e x a c t l y t h e$ $s c h e d u l e d t i m e, f i n d t h e n o r m a l s p e e d .$

Question Number 202058 Answers: 1 Comments: 1

Solve by computer programming a, b & c are Prime numbers. And they are in AP and d is common difference Example (a, b, c, d) = (3, 5, 7, 2) Just Next Set(a, b, c, d) = ?

$Solve by computer programming$ $a, b & c a r e P r i m e n u m b e r s . And they$ $are in AP and d is common difference$ $Example (a, b, c, d) = (3, 5, 7, 2)$ $Just Next Set (a, b, c, d) = ?$

Question Number 201952 Answers: 1 Comments: 0

(gof)_x =2x−1 (fog)_x ^(−1) =3x+2 (fof)_3 =?

${(g o f)}_{x} = 2 x - 1$ ${(f o g)}_{x}^{- 1} = 3 x + 2$ ${(f o f)}_{3} = ?$

Question Number 201850 Answers: 0 Comments: 0

Question Number 201433 Answers: 3 Comments: 0

A truck, P, travelling at 54km/h passes a point at 10:30 am while another truck, Q travelling at 90km/h passes through this same point 30 minutes later. At what time will truck Q overtake P?

$A t r u c k, P, t r a v e l l i n g a t 54 k m / h p a s s e s$ $a p o i n t a t 10 : 30 a m w h i l e a n o t h e r t r u c k,$ $Q t r a v e l l i n g a t 90 k m / h p a s s e s t h r o u g h$ $t h i s s a m e p o i n t 30 m i n u t e s l a t e r . A t$ $w h a t t i m e w i l l t r u c k Q o v e r t a k e P ?$

Question Number 201257 Answers: 2 Comments: 0

Biggest prime factor of (3^(14) + 3^(13) − 12) = ?

$Biggest prime factor of (3^{14} + 3^{13} - 12) = ?$

Question Number 201139 Answers: 1 Comments: 1

Question Number 201112 Answers: 1 Comments: 0

Question Number 200972 Answers: 1 Comments: 0

Question Number 200886 Answers: 4 Comments: 0

resoudre dans R : ((3+x))^(1/3) −((3−x))^(1/3) =((9−x^2 ))^(1/3)

$r e s o u d r e d a n s R :$ $\sqrt[3]{3 + x} - \sqrt[3]{3 - x} = \sqrt[3]{9 - x^{2}}$

Question Number 200657 Answers: 1 Comments: 0

Question Number 200627 Answers: 1 Comments: 0

Question Number 200531 Answers: 1 Comments: 0

Question Number 200321 Answers: 5 Comments: 0

for x,y,z ∈N, if 38x+40y+41z=520 find x+y+z=?

$f o r x, y, z \in N, i f$ $38 x + 40 y + 41 z = 520$ $f i n d x + y + z = ?$

Question Number 200265 Answers: 1 Comments: 0

Question Number 200236 Answers: 1 Comments: 0

what is the smallest natural number which has at least 100 divisors?

$w h a t i s t h e s m a l l e s t n a t u r a l n u m b e r$ $w h i c h h a s a t l e a s t 100 d i v i s o r s ?$

Question Number 200205 Answers: 0 Comments: 1

Question Number 200121 Answers: 0 Comments: 0

Question Number 200005 Answers: 3 Comments: 0

Question Number 200004 Answers: 1 Comments: 0

Question Number 199913 Answers: 0 Comments: 0

To calculate the mean and median of the distribution, you can follow these steps: 1. Calculate the midpoint for each weight range: - For 118-126: Midpoint = (118 + 126) / 2 = 122 - For 127-135: Midpoint = (127 + 135) / 2 = 131 - For 136-144: Midpoint = (136 + 144) / 2 = 140 - For 145-153: Midpoint = (145 + 153) / 2 = 149 - For 154-162: Midpoint = (154 + 162) / 2 = 158 - For 163-171: Midpoint = (163 + 171) / 2 = 167 - For 172-180: Midpoint = (172 + 180) / 2 = 176 2. Multiply each midpoint by its corresponding frequency to find the sum of the products. - (122 * 3) + (131 * 5) + (140 * 9) + (149 * 12) + (158 * 5) + (167 * 4) + (176 * 2) 3. Sum the frequencies to find the total number of data points. - 3 + 5 + 9 + 12 + 5 + 4 + 2 = 40 4. Calculate the mean (average) by dividing the sum of the products by the total number of data points: - Mean = (Sum of products) / (Total number of data points) 5. To find the median, you can start by listing the weights in ascending order. Then, locate the middle value. Since there are 40 data points (an even number), the median will be the average of the 20th and 21st values. - Arrange the midpoints in ascending order: 122, 131, 140, 149, 158, 167, 176. - The 20th value is 149, and the 21st value is 158. - Median = (149 + 158) / 2 Now, you can calculate the mean and median based on the given data.

Pg 1 Pg 2 Pg 3 Pg 4 Pg 5 Pg 6 Pg 7 Pg 8 Pg 9 Pg 10

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