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3 years ago

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Mathematics

1 answer:

Lana71 [14]3 years ago

6 0

In a triangle when an angle between two equal length sides is 60 that means that the triangle is equilateral.

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Solve for z: 2 + 8 - z = -24 Show your work

Gala2k [10]

2+8-z=-24
^
10-z=-24
-10 -10
——————-
-z=14
———
-z -z

Z=14

4 0

2 years ago

Please answer it now in two minutes

ziro4ka [17]

The correct answer is 4

Step by step is blew

Hopefully this help you

3 0

2 years ago

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A number cube is rolled 310 times the results are: 51 ones,56 twos,45 threes,54 fours,53 gives, and 51 sixes. What is the probab

kkurt [141]

A number cube is rolled 310 times the results are: 51 ones,56 twos,45 threes,54 fours,53 gives, and 51 sixes. What is the probability of rolling even number?
Ask for details Follow Report by Brianmugleston 32 seconds ago

4 0

3 years ago

Read 2 more answers

). If YU = YV, ST = 16, mQS = 34, and mRT = 98°, find each measure.

77julia77 [94]

Answer:

QR=8

QU=76

Measure of arc ST=114°

Measure of arc QR=114°

Measure of arc XT=57°

Step-by-step explanation:

Given that,

YU=YV

ST=16

mQS=34

mRT=98

We will use pythagoras theorem,

QR=8

QU=76

QR=ST

Since they span the same arc, let the arc be x.

360 - QS - RT = QR + ST

Recall that QR=ST

360 - 34 - 98 = x + x

228 = 2x

Divide both side by 2

X= 228/2

X=114°

Therefore, the measure of the arc ST and the arc QR=114°

Measure of span of arc XT will be:

= 1/2× (arc ST)

= 1/2 × 114

= 57°

8 0

3 years ago

In a church wing with 8 men and 10 women members find the probability that a 5 member committee chosen randomly will have.......

Sergeeva-Olga [200]

Answer:

a) Probability that a 5 member committee will have all men = 0.0065

b) probability that a 5 member committee chosen randomly will have 3 men and 2 women = 0.294

Step-by-step explanation:

Number of men = 8

Number of women = 10

Total number of members = 10 + 8 = 18

Probability = (Number of possible outcomes)/(Total number of outcomes)

Number of ways of selecting a 5 member committee from 18 people = $^{18}C_5 = \frac{18!}{(18-5)!5!} = \frac{18!}{13!5!}$ = 8568 ways

a) Probability that a 5 member committee will have all men

Number of ways of selecting 5 men from 8 men

= $^8C_5 = \frac{8!}{(8-5)!5!} = \frac{8!}{3!5!}$ = 56 ways

Probability that a 5 member committee will have all men = 56/8568

Probability that a 5 member committee will have all men = 0.0065

b)probability that a 5 member committee chosen randomly will have 3men and 2 women

Number of ways of selecting 3 men from 8 men

= $^8C_3 = \frac{8!}{(8-3)!3!} = \frac{8!}{5!3!}$ = 56 ways

Number of ways of selecting 2 women from 10 men

= $^{10}C_2 = \frac{10!}{(10-2)!2!} = \frac{10!}{8!2!}$ = 45 ways

Number of ways of selecting 3 men and 2 women = 56*45

Number of ways of selecting 3 men and 2 women = 2520

Probability of selecting 3 men and 2 women = 2520/8568 = 0.294

probability that a 5 member committee chosen randomly will have 3 men and 2 women = 0.294

6 0

3 years ago