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Hoochie [10]
3 years ago
13

Ben is 4 times as old as Ishaan and is 6 years older than ishaan. How old is ben?

Mathematics
2 answers:
ale4655 [162]3 years ago
7 0

Answer:

8 years old

Step-by-step explanation:

masya89 [10]3 years ago
6 0

Answer:

1.5 years old

Step-by-step explanation:

1.5 times 4 is 6

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Honey and Money are twin brothers. Mother wants to equally divide a rectangular shaped cake of 5cm wide and 6cm long. Find the l
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Answer:

Length of rectangular cake = 5 cm

Breadth of rectangular cake = 6 cm

Area of rectangle = 5 cm × 6 cm =30 cm²

If cake is equally distributed between Honey and Money, then areas of two parts should be equal.

So, Amount of cake that Honey and money got \frac{30}{2}=15

Share of each (Honey and Money)=15 cm²

So, if the third cake if it was cut diagonally ,the cake must have some height.

Let the height of cake be H.

So, Height of third cake =\sqrt{L^2+B^2+H^2}\\\\=\sqrt{5^2+6^2+H^2}\\\\=\sqrt{61+H^2}

If , H=0 that is height is negligible having 0 thickness,

Then the Length of third cake, which is divided into two parts=√61=7.81 cm

5 0
3 years ago
Six cards numbered from 1 to 6 are placed in an empty bowl. First one card is drawn and then put back into the bowl; then a seco
kari74 [83]

Answer:

C. \frac{1}{18}

Step-by-step explanation:

Given: Six cards numbered from 1 to 6 are placed in an empty bowl. First one card is drawn and then put back into the bowl then a second card is drawn.

To Find: If the cards are drawn at random and if the sum of the numbers on the cards is 8, what is the probability that one of the two cards drawn is numbered 5.

Solution:

Sample space for sum of cards when two cards are drawn at random is \{(1,1),(1,2),(1,3)......(6,6)\}

total number of possible cases =36

Sample space when sum of cards is 8 is \{(3,5),(5,3),(6,2),(2,6),(4,4)\}

Total number of possible cases =5

Sample space when one of the cards is 5 is \{(5,3),(3,5)\}

Total number of possible cases =2

Let A be the event that sum of cards is 8

p(\text{A}) =\frac{\text{total cases when sum of cards is 8}}{\text{all possible cases}}

p(\text{A})=\frac{5}{36}

Let B be the event when one of the two cards is 5

probability than one of two cards is 5 when sum of cards is 8

p(\frac{\text{B}}{\text{A}})=\frac{\text{total case when one of the number is 5}}{\text{total case when sum is 8}}

p(\frac{\text{B}}{\text{A}})=\frac{2}{5}

Now,

probability that sum of cards 8 is and one of cards is 5

p(\text{A and B}=p(\text{A})\times p(\frac{\text{B}}{\text{A}})

p(\text{A and B})=\frac{5}{36}\times\frac{2}{5}

p(\text{A and B})=\frac{1}{18}

if sum of cards is 8 then probability that one of the cards is 8 is \frac{1}{18}, option C is correct.

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The function f(x) varies inversely with x and f(x)=0.5 when x = 0.3. What is f(x) when x = 2.4? A)5.76 B)0.0625 C)0.30 D)0.15
goldfiish [28.3K]

Answer:

B) 0.0625

Step-by-step explanation:

f(x) = k / x

f(x)=0.5 when x = 0.3

f(x) = k / x

0.5 = k / 0.3

Cross product

0.5 * 0.3 = k

k = 0.15

What is f(x) when x = 2.4?

f(x) = k / x

f(x) = 0.15 / 2.4

f(x) = 0.0625

The answer is B) 0.0625

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