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jeyben [28]
2 years ago
10

36. How many 13 -card hands dealt from a standard deck will have exactly seven spades?

Mathematics
1 answer:
barxatty [35]2 years ago
7 0

Answer:

id k give a second

Step-by-step explanation:

You might be interested in
What is bigger 3/10 or 3/8
olasank [31]
3/8 is bigger than 3/10. In maths 10ths are smaller than 8ths. If both have three parts it means that 3/8 must be bigger
7 0
3 years ago
Read 2 more answers
What is. - 3 7/8 × - 4 3/5 help please
Yakvenalex [24]

Answer:

713/40 or as mixed fraction of 17 33/40 (see setps below)

Step-by-step explanation:

Simplify the following:

-(3 + 7/8) (-(4 + 3/5))

Hint: | Multiply all instances of -1 in -(3 + 7/8) (-(4 + 3/5)).

(-1)^2 = 1:

(4 + 3/5) (3 + 7/8)

Hint: | Put the fractions in 3 + 7/8 over a common denominator.

Put 3 + 7/8 over the common denominator 8. 3 + 7/8 = (8×3)/8 + 7/8:

(4 + 3/5) (8×3)/8 + 7/8

Hint: | Multiply 8 and 3 together.

8×3 = 24:

(4 + 3/5) (24/8 + 7/8)

Hint: | Add the fractions over a common denominator to a single fraction.

24/8 + 7/8 = (24 + 7)/8:

(4 + 3/5) (24 + 7)/8

Hint: | Evaluate 24 + 7 using long addition.

| 1 |  

| 2 | 4

+ | | 7

| 3 | 1:

(4 + 3/5)×31/8

Hint: | Put the fractions in 4 + 3/5 over a common denominator.

Put 4 + 3/5 over the common denominator 5. 4 + 3/5 = (5×4)/5 + 3/5:

31/8 (5×4)/5 + 3/5

Hint: | Multiply 5 and 4 together.

5×4 = 20:

((20/5 + 3/5)×31)/8

Hint: | Add the fractions over a common denominator to a single fraction.

20/5 + 3/5 = (20 + 3)/5:

31/8 (20 + 3)/5

Hint: | Evaluate 20 + 3.

20 + 3 = 23:

23/5×31/8

Hint: | Express 23/5×31/8 as a single fraction.

23/5×31/8 = (23×31)/(5×8):

(23×31)/(5×8)

Hint: | Multiply 5 and 8 together.

5×8 = 40:

(23×31)/40

Hint: | Multiply 23 and 31 together.

| 3 | 1

× | 2 | 3

| 9 | 3

6 | 2 | 0

7 | 1 | 3:

Answer: 713/40

________________________________________

Convert to a mixed number:

713/40

Hint: | Use long division to get a quotient and remainder.

Divide 713 by 40:

4 | 0 | 7 | 1 | 3

Hint: | How many times does 40 go into 71?

40 goes into 71 at most one time:

| | | 1 | |  

4 | 0 | 7 | 1 | 3 |  

| - | 4 | 0 | |  

| | 3 | 1 | 3 |  

Hint: | How many times does 40 go into 313?

40 goes into 313 at most 7 times:

| | | 1 | 7 |  

4 | 0 | 7 | 1 | 3 |  

| - | 4 | 0 | |  

| | 3 | 1 | 3 |  

| - | 2 | 8 | 0 |  

| | | 3 | 3 |  

Hint: | Summarize the results.

Read off the results. The quotient is the number at the top and the remainder is the number at the bottom:

| | | 1 | 7 | (quotient)

4 | 0 | 7 | 1 | 3 |  

| - | 4 | 0 | |  

| | 3 | 1 | 3 |  

| - | 2 | 8 | 0 |  

| | | 3 | 3 | (remainder)

Hint: | In the equivalent mixed fraction, the quotient becomes the whole number part, the remainder becomes the numerator part, and the divisor (denominator) remains the denominator.

The quotient of 713/40 is 17 with remainder 33, so:

Answer: 17 33/40

6 0
3 years ago
A probability experiment is conducted in which the sample space of the experiment is upper s equals startset 4 comma 5 comma 6 c
olchik [2.2K]
The correct question statement is:

A probability experiment is conducted in which the sample space of the experiment is S = {4,5,6,7,8,9,10,11,12,13,14,15}. Let event E={7,8,9,10,11,12,13,14,15}. Assume each outcome is equally likely. List the outcomes in E^{c}. Find P(E^{c}).

Solution:

Part 1:

E^{c} means compliment of the set E. A compliment of a set can be obtained by finding the difference of the set from the universal set. The universal set is the set which contains all the possible outcomes of the events which is S in this case.

So, compliment of E will be equal to S - E. S - E will result in all those elements of S which are not present in E. So, we can write:

E^{c}=S-E \\  \\ 
E^{c}=(4,5,6,7,8,9,10,11,12,13,14,15)-(7,8,9,10,11,12,13,14,15) \\  \\ 
E^{c}=(4,5,6)

Thus the set compliment of E will contain the elements {4,5,6}.So

E^{c} = {4,5,6}

Part 2)

P(E^{c}) means probability that if we select any number from the Sample Space S, it will belong the set E compliment.

P(E^{c}) = (Number of Elements in E^{c})/Number of elements in S

Number of elements in set S = n(S) = 12
Number of elements in set E^{c} = n(E^{c})=3

So, 

P(E^{c})= \frac{n(E^{c}) }{n(S)} \\  \\ 
P(E^{c})= \frac{3}{12} \\  \\ 
P(E^{c})= \frac{1}{4}
7 0
3 years ago
An empty bucket weighs 800g.
erica [24]

Answer:

412.5 percent

7 0
3 years ago
Read 2 more answers
Solve for t. 5(t + 3)= -3.5
sattari [20]

Answer:

Step-by-step explanation:

5(t+3)=-3.5

distribute

5t+15=-3.5

subtract 15 from both sides

5t= -18.5

divide both sides by 5

t=-3.7

5 0
2 years ago
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