Use the diameter if their is no radius.
Side of the original square. x2 x<span>. Square </span>has all<span> sides the same </span>length x<span> ... </span>2<span>. Our Solution. Example 3. The </span>length<span> of a </span>rectangle<span> is </span>4<span> ft greater than the </span>width<span>. If each </span>dimension<span> is ... (x+ 3)(x+ 7) =</span>x<span>(x+ </span>4<span>) +33 Set up equation, </span>length times width<span> is area .... </span>5<span>) </span>6 x 10<span>. </span>6<span>) 20 ft </span>x<span> 35 ft. 7) </span>6<span>” </span>x 6<span>”. </span>8<span>) </span>6<span> yd </span>x<span> 7 yd. </span>9<span>) </span>4<span> ft </span>x<span> 12 ft. </span>10<span>) 1.54 in.</span>
Hi
7√8*4√5
= 56√10
I hope that's help and sorry for late answer ^...^
Answer: 4.1
Step-by-step explanation:
Given : An urn contains 10 balls with the number ‘5’ printed on them, 6 balls with ‘4’ printed on them, and 4 balls with ‘2’ printed on them.
Total balls = 10+6+4=20
Let A , B and C are the events of drawing ball with the number ‘5’, ball with the number ‘4’ and ball with the number ‘2’ respectively.
Then,
If X is the number observed.
Since ![E[X]=\sum_{i=1}^{n}x_ip_i](https://tex.z-dn.net/?f=E%5BX%5D%3D%5Csum_%7Bi%3D1%7D%5E%7Bn%7Dx_ip_i)
Then,
![E[X]=5\times P(A)+4\times P(B)+2\times P(C)\\\\=5\times 0.5+4\times 0.3+2\times 0.2\\=4.1](https://tex.z-dn.net/?f=E%5BX%5D%3D5%5Ctimes%20P%28A%29%2B4%5Ctimes%20P%28B%29%2B2%5Ctimes%20P%28C%29%5C%5C%5C%5C%3D5%5Ctimes%200.5%2B4%5Ctimes%200.3%2B2%5Ctimes%200.2%5C%5C%3D4.1)
Hence, E[X]=4.1
