A. there are 72 right-handed batters. this is what you do:
5:8
45:?
45/5=9
8*9=72 so
45:72
b. there are 45 left-handed batters
the total batters is 45+72=117
so the ratio of left-handed to the total batter would be 45:117
hope this helps :)

4 0

3 years ago

Twenty-six , 2 tens 6 ones , 25, 20+6 which one is the odd one out?

Aneli [31]

Answer:

Step-by-step explanation:

twenty-six=26

2 tens 6 ones= 26

25=25

20+6=26

25 is the odd one out.

8 0

3 years ago

What + 89= 9(6+5) please solve for my answer and explain how you got your answer

nlexa [21]

The answer would be 10.

6 + 5 = 11
11 x 9 = 99

This means 9(6+5) = 99

99 - 89 = 10

6 0

3 years ago

An insurance company divides its policyholders into low-risk and high-risk classes. 60% were in the low-risk class and 40% in th

AleksAgata [21]

Answer:

a). 0.294

b) 0.11

Step-by-step explanation:

From the given information:

the probability of the low risk = 0.60

the probability of the high risk = 0.40

let $C_o$ represent no claim

let $C_1$ represent 1 claim

let $C_2$ represent 2 claim :

For low risk;

so, $C_o$ = (0.80 * 0.60 = 0.48), $C_1$ = (0.15* 0.60=0.09), $C_2$ = (0.05 * 0.60=0.03)

For high risk:

$C_o$ = (0.50 * 0.40 = 0.2), $C_1$ = (0.30 * 0.40 = 0.12) , $C_2$ = ( 0.20 * 0.40 = 0.08)

Therefore:

a), the probability that a randomly selected policyholder is high-risk and filed no claims can be computed as:

$P(H|C_o) = \dfrac{P(H \cap C_o)}{P(C_o)}$

$P(H|C_o) = \dfrac{(0.2)}{(0.48+0.2)}$

$P(H|C_o) = \dfrac{(0.2)}{(0.68)}$

$P(H|C_o) = 0.294$

b) What is the probability that a randomly selected policyholder filed two claims?

the probability that a randomly selected policyholder be filled with two claims = 0.03 + 0.08

= 0.11

7 0

3 years ago