In ten-pin bowling, the highest possible score in a single game is 300. At one point in the bowling season, Fred F Stone had an average score of 177. In his next game he obtained a score of 199, which caused his average to increase to 178. After one more game Fred would like his average to be 183.
Is it possible for Fred to accomplish this? If it is possible, what score does he need in his next game? If it is not possible, explain why not.

a) Is it possible for Fred to accomplish this? Yes

b) If it is possible, what score does he need in his next game? 293

Step-by-step explanation:

Step 1

Let us represent the number of games Fred bowled to get an average of 177 = X

His total points scored in that game would be

177 × X

= 177X

To achieve an average of 178 for his next game,

The total number of points he scored was 199, hence that is represented as:

178 = 177X + 199/ X + 1

178(X + 1) = 177X + 199

178X + 178 = 177X + 199

178X - 177X = 199 - 178

X = 21

Hence, Fred bowled 21 games to achieve an average of 178

Therefore, the score he needs to have on the 23rd game is obtained as:

= (23 × 183) - (22 × 178) = 4209 - 3916 = 293

Therefore, it is possible for Fred to raise his average from 178 to 183 in a single game, but he must bowl 293 in his next game to do this.

< xcy ≅ < xby

step-by-step explanation:

in δcax and δbax,

so, δcax ≅ δbax (side-angle-side or sas)

since δcax ≅ δbax, we can conclude < acx ≅ < abx

in δacy and δaby,

so, δacy ≅ δaby (side-angle-side or sas)

since δacy ≅ δaby, we can conclude < acy ≅ < aby

now,

< acy ≅ < aby

=> < acx + < xcy ≅ < abx + < xby

=> < acx + < xcy ≅ < acx + < xby   (since < acx ≅ < abx already proved above)

subtracting < acx from both sides, we get

< acx + < xcy -< acx ≅ < acx + < xby -< acx

cancelling out < acx and -< acx from both sides, we get

< xcy ≅ < xby (proved)