All categories

3 years ago

What is the long division work of the problem of 337 divided by 2.

Mathematics

2 answers:

padilas [110]3 years ago

5 0

Answer:

168.5

Step-by-step explanation:

₁ ₁ ₁

2 ) 3 3 7. 0

1 6 8.5 <--- Quoitient

first part is 2 divide once into 3 with remainder 1

- and this 1 goes in front of the 3 to give 13.

2 into 13 goes 6 times with remainder 1

- then we get 2 into 17 giving 8 remainder 1 and so on....

MatroZZZ [7]3 years ago

4 0

Step-by-step explanation:

168 <-- quotient --- 337 2 - 13 12 -- 17 16 -- 1 <-- remainder

You might be interested in

Plz answer will mark brainliest

LUCKY_DIMON [66]

Answer:

(5.5, 6)

Step-by-step explanation:

Find the difference between them and then divide by 2 and add to one point

7 0

3 years ago

The radius of a circle is 14 meters. What is the circle circumference

seropon [69]

Step-by-step explanation:

HERE,

redius of a circle (r)=14 meters.

we know that,

$\boxed{\red{\sf{circle~ circumference=2\times \pi \times r }}}$

According to the question, 

circle circumference=

$\tt{2\times \pi \times 14 }$

$\tt{2 \times \dfrac{22}{7} \times 14 }$

$\tt{2 \times \dfrac{22}{\cancel{7}} \times \cancel{14}}$

$\tt{2\times 22 \times 2 }$

$\tt{ 88~meters }$

so, the circumference of that circle is 88 meters.

5 0

3 years ago

A tree casts a shadow x = 100 ft long when a vertical rod 6.0 ft high casts a shadow 4.0 ft long. (See the illustration.) How ta

rewona [7]

I'm not really sure If I would be much help but I'll try best k? (cries a little) So you do 100
+ 6.0
+ 4.0
_________________
11 .0

So your answer is 11.0 for the tall tree.

8 0

3 years ago

(3 points) Blades of grass Suppose that the heights of blades of grass are Normally distributed and independent, with each heigh

NemiM [27]

The final answer is:

a) P( Y < 42.5 ) = 0.8541

b) P( 39.5 < Y < 40.5 ) = 0.1670.

What is the normal distribution?

A continuous probability distribution for a real-valued random variable in statistics is known as a normal distribution or Gaussian distribution.

If x follows a normal distribution with mean μ and standard deviation σ then the distribution of

$\sum_{i =1}^{n}x_{i}$ follows an approximately normal distribution with a mean $n\mu$ and standard deviation $\sqrt{n }\sigma$