Answer:
8. x = 16
9. x = 10
14.
m ∠RSU = 130°
m ∠UST = 50°
15.
m ∠RSU = 124°
m ∠UST = 56°
Step-by-step explanation:
8.
Given ∠DEF is bisected by EG. That is , ∠DEG = ∠GEF
That is , (x + 15)° = 31°
x = 31 - 15 = 16
9.
Given ∠DEF is bisected by EG. That is , ∠DEG = ∠GEF
That is ,
(6x - 4)° = 56°
6x = 56 + 4
6x = 60
x = 10
14.
13x + 5x = 180° [straight line angles ]
18x = 180
x = 10
m ∠RSU = 130°
m ∠UST = 50°
15.
4x + 12 + 2x = 180° [ straight line angles]
6x = 180 - 12
6x = 168
x = 28
m ∠RSU = 4(28) + 12 = 112 + 12 = 124°
m ∠UST = 2(28) = 56°
Let X be the first and second number.
X+X = 18
+ X-X = 2
2X = 20 (divide both sides by 2
X = 10
Substitute the values to get the second number
10+X = 18
X = 18 -10
X = 8
Therefore, the numbers are 10 and 8
106 because, 85 plus 21 = 106
Answer:
The probability that in a one-game playoff, her score is more than 227 is 0.4404.
Step-by-step explanation:
We are given that Susan has been on a bowling team for 14 years. After examining all of her scores over that period of time, she finds that they follow a normal distribution. Her average score is 225, with a standard deviation of 13.
<em>Let, X = scores over that period of time</em>
X ~ N(
)
The z score probability distribution is given by;
Z =
~ N(0,1)
where,
= average score
= standard deviation
So, probability that in a one-game playoff, her score is more than 227 is given by = P(X > 227)
P(X > 227) = P(
<
) = P(Z > 0.15) = 1 - P(Z
0.15)
= 1 - 0.55962 = 0.4404
Therefore, probability that in a one-game playoff, her score is more than 227 is 0.4404.
Step-by-step explanation:
0.002
ones is 0
tenths is 0
hundreths is 0
thousandths is 2