Ask question

Ask question

All categories

dimulka [17.4K]

4 years ago

12

The ratio of the number of left handers to the number of right handers in a middle school is 6:15. If there are 120 left handers

how many right handers are there?

1 answer:

Zinaida [17]4 years ago

7 0

Answer:300

Step-by-step explanation:

there are 9 15's in 120 which means its going by 15's 3 x 5= 15 and if it says 6 that means its timesed by 2. 15 x 2= 30 then 30 x 10= 300

You might be interested in

What's two thirds plus six seventh?

vfiekz [6]

Answer:

32/21

Step-by-step explanation:

multiply (2/3) by (7/7) and (6/7) by (3/3). (14/21) + (18/21) is (32/21)

7 0

3 years ago

Read 2 more answers

Karen had 3 apples. she ate 2. how many will she have left

Vadim26 [7]

Answer:1

Step-by-step explanation:

7 0

3 years ago

If 5/8 of the people in water aerobics class are over age 65 and 1/4 of the people in the class are under age 40,what fraction o

LiRa [457]

I think the answer's 7/8. If you convert 1/4 to 2/8 and add them together, you add 5/8 and 2/8 to get 7/8.

6 0

3 years ago

The triangles below are similar. Triangle A B C. Side A C is 10 and side A B is 5. Angle C is 30 degrees. Triangle D E F. Side E

Gnoma [55]

Answer:

$\triangle ABC {\displaystyle \sim } \triangle DE\ F$

$\triangle CBA {\displaystyle \sim } \triangle FED$

$\triangle BAC {\displaystyle \sim } \triangle EDF$

Step-by-step explanation:

Given

$\triangle ABC {\displaystyle \sim } \triangle DE\ F$

$AC = 10$

$AB = 5$

$\angle C = 30^\circ$

$ED = 7.5$

$DF = 25$

$\angle F =30$

$\angle E =90$

Required

Which of the options is/are true:

$\triangle CBA {\displaystyle \sim } \triangle FED$ $\triangle CBA {\displaystyle \sim } \triangle FDE$ $\triangle BAC {\displaystyle \sim } \triangle EFD$

$\triangle BAC {\displaystyle \sim } \triangle EDF$ $\triangle ABC {\displaystyle \sim } \triangle DE\ F$ $\triangle ABC {\displaystyle \sim } \triangle DFE$

The given triangles, implies that:

$A {\displaystyle \sim } \ D$

$B {\displaystyle \sim } \ E$

$C {\displaystyle \sim } \ F$

By taking each sides of both triangle, one after the other; the possible similar triangles are:

$\triangle ABC {\displaystyle \sim } \triangle DE\ F$

$\triangle CBA {\displaystyle \sim } \triangle FED$

$\triangle BAC {\displaystyle \sim } \triangle EDF$

5 0

3 years ago

Helppppppppppppppppp

Komok [63]

The answer would be c

6 0

4 years ago