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Alex_Xolod [135]

3 years ago

6

Oihuivouvogcopvcbotfo7uf7uyvluyci76yiuiyfiytulyfvi;gougcirytfcvy;ivp u;tvpi;uolgvoulyik8,tgovuyjfugohl8iukgvb kycvi yii5rfteusxd

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1 answer:

Helga [31]3 years ago

6 0

Answer:

i would love to help

Step-by-step explanation:

but i dont get what this is =-=

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The measure of one interior angle is 180 degrees find the number of sides

Masteriza [31]

If it has 3-2 sides it is Regular
If it has 4-2 sides it is a 4 sided polygon or a (quadrilateral)
Lastly 6-2 is a 6 sided polygon or a (hexagon)

5 0

3 years ago

Six cards numbered from 1 to 6 are placed in an empty bowl. First one card is drawn and then put back into the bowl; then a seco

kari74 [83]

Answer:

C. $\frac{1}{18}$

Step-by-step explanation:

Given: Six cards numbered from $1$ to $6$ are placed in an empty bowl. First one card is drawn and then put back into the bowl then a second card is drawn.

To Find: If the cards are drawn at random and if the sum of the numbers on the cards is $8$ , what is the probability that one of the two cards drawn is numbered $5$ .

Solution:

Sample space for sum of cards when two cards are drawn at random is $\{(1,1),(1,2),(1,3)......(6,6)\}$

total number of possible cases $=36$

Sample space when sum of cards is $8$ is $\{(3,5),(5,3),(6,2),(2,6),(4,4)\}$

Total number of possible cases $=5$

Sample space when one of the cards is $5$ is $\{(5,3),(3,5)\}$

Total number of possible cases $=2$

Let A be the event that sum of cards is $8$

$p(\text{A})$ $=\frac{\text{total cases when sum of cards is 8}}{\text{all possible cases}}$

$p(\text{A})=\frac{5}{36}$

Let B be the event when one of the two cards is $5$

probability than one of two cards is $5$ when sum of cards is $8$

$p(\frac{\text{B}}{\text{A}})=\frac{\text{total case when one of the number is 5}}{\text{total case when sum is 8}}$

$p(\frac{\text{B}}{\text{A}})=\frac{2}{5}$

Now,

probability that sum of cards $8$ is and one of cards is $5$

$p(\text{A and B}=p(\text{A})\times p(\frac{\text{B}}{\text{A}})$

$p(\text{A and B})=\frac{5}{36}\times\frac{2}{5}$

$p(\text{A and B})=\frac{1}{18}$

if sum of cards is $8$ then probability that one of the cards is $8$ is $\frac{1}{18}$ , option C is correct.

3 0

3 years ago

Kristin wants to ask the managers of the fast food restaurants in a large city “How do you retain employees?” What condition mus

Pani-rosa [81]

Answer:

d

Step-by-step explanation:

3 0

3 years ago

If f(y) = 2y - 2/5,<br> what is f(f^-1(1024))?

SCORPION-xisa [38]

Answer:

-0.399

Step-by-step explanation:

f^-1(1024)=(2*1024-2/5)^-1= 0.000488

f(0.000488)=2*0.000488-2/5= -0.399

7 0

3 years ago

Help me out , plzzzzz!!

natta225 [31]

Answer:

a. 5,040 ways

b. 210 ways

Step-by-step explanation:

a) We want to pick 4 numbers out of 10 given that orders matter

If orders do matter, it will be a permutation problem

Mathematically;

n P r = n!/(n-r)!

In this case, n is 10 and r is 4

Thus, we have it that;

10 P 4 = 10!/(10-4)! = 5,040

b) if orders do not matter

It will be a combination problem

n C r = n!/(n-r)!r!

n = 10 and r = 4

10 C 4 = 10! /(10-4)!4!

= 210

3 0

3 years ago