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Mazyrski [523]
3 years ago
10

What is2/3 divided by 4/5

Mathematics
2 answers:
Sloan [31]3 years ago
8 0
The answer to this question is 5/6
OLEGan [10]3 years ago
3 0

Step-by-step explanation:

<u>2</u><u> </u>/<u>4</u>

3. 5

doing reciprocal

<u>2</u>× <u>5</u>

3. 4

= <u>8</u><u>×</u><u>1</u><u>5</u>

<u> </u><u> </u><u> </u><u> </u><u> </u><u>1</u><u>2</u>

= <u>1</u><u>2</u><u>0</u>

<u> </u><u> </u><u> </u><u> </u><u> </u>12

=10 ans

hope it help u..

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If a non zero number is divided by a zero then what is the notation to denote it​
enot [183]
Answer:

0 Divided by a Number 0a=0 Dividing 0 by any number gives us a zero. Zero will never change when multiplying or dividing any number by it.

if i helped please give brainliest and a thanks !!
3 0
3 years ago
А square tile has a width of foot. How many tiles will fit end-to-end along a<br> 4-foot wall?
irina1246 [14]

Answer:

4 tiles

Step-by-step explanation:

4/1 = 4

5 0
2 years ago
Read 2 more answers
Ship collisions in the Houston Ship Channel are rare. Suppose the number of collisions are Poisson distributed, with a mean of 1
alexandr1967 [171]

Answer:

a) \simeq 0.3012   b) \simeq 0.0494 c) \simeq 0.2438

Step-by-step explanation:

Rate of collision,

1.2 collisions every 4 months

or, \frac{1.2}{4}

= 0.3 collisions per  month

So, the Poisson distribution for the random variable no. of collisions per month (X) is given by,

          P(X =x) = \frac{e^{-\lambda}\times {\lambda}^{x}}}{x!}&#10;

                                                           for x ∈ N ∪ {0}

                       =  0 otherwise --------------------------------------(1)

here, \lambda = 0.3 collision / month

No collision over a 4 month period means no collision per month or X =0

Putting X = 0 in (1) we get,

         P(X = 0) = \frac{e^{-0.3}\times {\0.3}^{0}}{0!}&#10;

                      \simeq 0.7408182207 ------------------------------------(2)

Now, since we are calculating  this for 4 months,

so, P(No collision in 4 month period)

     =0.7408182207^{4}

     \simeq 0.3012  -----------------------------------------------------------(3)

2 collision in 2 month period means 1 collision per month or X =1

Putting X =1 in (1) we get,

           P(X =1) = \frac{e^{-0.3}\times {\0.3}^{1}}{1!}&#10;

                      \simeq 0.2222454662 ------------------------------------(4)

Now, since we are calculating this for 2 months, so ,

P(2 collisions in 2 month period)

                =0.2222454662^{2}

                \simeq 0.0494 -----------------------------------------(5)

1 collision in 6 months period means

                                \frac{1}{6} collision per month

Now, P(1 collision in 6 months period)

= P( X = 1/6]  (which is to be estimated)

=\frac {P(X=0)\times 5 + P(X =1)\times 1}{6}

= \frac {0.7408182207 \times 5 + 0.2222454662 \times 1}{6}[/tex]

\simeq 0.6543894283-------------------------------------------(6)

So,

P(1  collision in 6 month period)

  =  0.6543894283^{6}

   \simeq 0.0785267444 ------------------------------------------------(7)

So,

P(No collision in 6 months period)

  = (P(X =0)^{6}

   \simeq 0.1652988882 ---------------------------------(8)

so,

P(1 or fewer collision in 6 months period)

= (8) + (7 ) = 0.0785267444 +0.1652988882

\simeq  0.2438 ---------------------------------------------(9)          

7 0
3 years ago
Can u pls help me with this question ​
Shalnov [3]

Answer:

A, C, D

Step-by-step explanation:

7 0
2 years ago
Read 2 more answers
The sector of a circle with a 60 millimeters radius has a central angle measure of 30° . What is the exact area of the sector in
ASHA 777 [7]
R = 60
60^2 * pi = 3600 * pi
angle = 30°
3600pi / 30 = 120pi
the answer is 120 * pi
7 0
3 years ago
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