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3 years ago

Perform the indicated operation. 61/2 · 4/5? a.5 1/5 b.6 2/5 c.3 1/5 d.4 2/5

Mathematics

2 answers:

dexar [7]3 years ago

5 0

Answer:

5 1/5

Step-by-step explanation:

6 1/2 *4/5

Rewriting as an improper fraction

( 2*6 +1)/2 *4/5

13/2 *4/5

52/ 10

50/10 + 2/10

5 +1/5

5 1/5

Bas_tet [7]3 years ago

4 0

Answer:

A. 5 1/5

Step-by-step explanation:

6 1/2 × 4/5

Convert 6 1/2 to an improper fraction.

13/2 × 4/5

Multiply both fractions.

52/10

Convert the improper fraction to a mixed fraction.

5 2/10

⇒ 5 1/5

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Rita is making a beaded bracelet. She has a collection of 160 blue beads, 80 gray beads, and 240 pink beads. What is the estimat

a_sh-v [17]

Step-by-step explanation:

Given that Rita is making a beaded bracelet. She has a collection of 160 blue beads, 80 gray beads, and 240 pink beads. We are to calculate the probability that Rita will need to pick atleast 5 beads before she picks a grey bead from her collection.

Prob for drawing atleast 5 beads before she picks a grey bead from her collection

= 1-Prob for drawing atleast one grey beed in the first 5 draws.

(Because these two are complementary events)

no of grey beeds drawn in first 5 trials is

Bi=(5,1/6)

Prob for drawing atleast one grey beed in the first 5 draws.

=1-Prob of no grey

Hence required prob=P(X=0 in first 5 draws)

= 0.4018

6th beeds onwards can be grey also.

Nearest answer is c)0.45

7 0

3 years ago

First right answer for both gets brain

kirza4 [7]

B 8:00 am

8:00+1:25=9:25
9:25+25=9:50
9:50+30=10:20

B $240

60×2=120×2=240

7 0

2 years ago

) An instructor gives his class a set of 18 problems with the information that the next quiz will consist of a random selection

RSB [31]

Answer:

The probability the he or she will answer correctly is 1.5%

Step-by-step explanation:

In all, there are 18 problems. In this question, the order of which the problems are sorted for the quiz makes no difference. For example, if the question A of the quiz is P1 and question B P2, and question A P2 and question B P1, it is the same thing.

There are 18 problems and 9 are going to be selected. So, there is going to be a combination of 9 elements from a set of 18 elements.

A combination of n elements from a set of m objects has the following formula:

$C_{(m,n)} = \frac{m!}{n!(m-n)!}$

In this question, m = 18, n = 9. So the total number of possibilities is:

$T_{p} = C_{(18,9)} = \frac{18!}{9!(18-9)!} = 48620$

Now we have to calculate the number of desired outcomes. This number is a combination of 9 elements from a set of 13 elements(13 is the number of problems that the student has figured out how to do).

Now, m = 13, n = 9. The number of desired possibilities is:

$D_{p} = C_{(13,9)} = \frac{13!}{9!(13-9)!} = 715$