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3 years ago

Estimate the quotient and then solve 13.2 ÷ 6 = ________.

Mathematics

2 answers:

andrezito [222]3 years ago

8 0

Answer:

2.2 is the answer

Step-by-step explanation:

Oksanka [162]3 years ago

4 0

Answer:

2.2

Step-by-step explanation:

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Suppose that the universal set is U={1,2,3,4,5,6,7,8,9,10}. Express each of the following subsets with bit strings (of length 10

Vladimir [108]

Answer:

0011100000

1010010001

0111001110

Step-by-step explanation:

As the question is not complete, Here is the complete question.

Suppose that the universal set is U = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}. Express each of these sets with bit strings where the ith bit in the string is 1 if i is in the set and 0 otherwise.

a) {3, 4, 5}

b) {1, 3, 6, 10}

c) {2, 3, 4, 7, 8, 9}.

So, we need to express a) b) and c) into bit strings.

Firstly, number of elements in the universal set represent the number of bits in the bit string.

Secondly, 1 = yes element is present in both universal set as well as in sub set.

0 = No, element is not present in sub set but present in universal set.

Hence, we have:

a) Sub set {3,4,5} = 0011100000 (As there are 3 1's which means only 3,4,5 are present in both universal set and subset.

Similarly,

b) Sub set {1, 3, 6, 10} = 1010010001

c) Sub set {2, 3, 4, 7, 8, 9} = 0111001110

5 0

3 years ago

Find the solution set

AVprozaik [17]

M1= -5 , m2 = 5 inorder to get this you move the constant to the right take the root of both sides separate the solutions and you should get m1= -5, m2 =5

7 0

3 years ago

Please help, thank you

alexandr1967 [171]

Answer:

falseeeeeeeee for sure

Step-by-step explanation:

can i get brianlissssstttt

4 0

3 years ago

The probability that a professor arrives on time is 0.8 and the probability that a student arrives on time is 0.6. Assuming thes

saul85 [17]

Answer:

a)0.08 , b)0.4 , C) i)0.84 , ii)0.56

Step-by-step explanation:

Given data

P(A) = professor arrives on time

P(A) = 0.8

P(B) = Student aarive on time

P(B) = 0.6

According to the question A & B are Independent

P(A∩B) = P(A) . P(B)

Therefore

${A}'$ & ${B}'$ is also independent

${A}'$ = 1-0.8 = 0.2

${B}'$ = 1-0.6 = 0.4

part a)

Probability of both student and the professor are late

P(A'∩B') = P(A') . P(B') (only for independent cases)

= 0.2 x 0.4

= 0.08

Part b)

The probability that the student is late given that the professor is on time

$P(\frac{B'}{A})$ = $\frac{P(B'\cap A)}{P(A)}$ = $\frac{0.4\times 0.8}{0.8}$ = 0.4

Part c)

Assume the events are not independent

Given Data

P $(\frac{{A}'}{{B}'})$ = 0.4

= $\frac{P({A}'\cap {B}')}{P({B}')}$ = 0.4

$P$ $({A}'\cap {B}')$ = 0.4 x P $({B}')$

= 0.4 x 0.4 = 0.16

$P({A}'\cap {B}')$ = 0.16

The probability that at least one of them is on time

$P(A\cup B)$ = 1- $P({A}'\cap {B}')$

= 1 - 0.16 = 0.84

ii)The probability that they are both on time

P $(A\cap B)$ = 1 - $P({A}'\cup {B}')$ = 1 - $[P({A}')+P({B}') - P({A}'\cap {B}')]$

= 1 - [0.2+0.4-0.16] = 1-0.44 = 0.56

6 0

3 years ago

There are 10 men and 8 women who want to be a part of a commitee of 5 people. If 2 women and 3 men are chosen, how many possible

icang [17]

Their can be 6 arrangements of (2 men to
3 women, and opposite 5 men to 0 women and opposite, 4 men and 1 woman and opposite. so therefore their are 6 arrangements

4 0

3 years ago