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3 years ago

Question 7 of 27

Mathematics

1 answer:

Alisiya [41]3 years ago

5 0

Answer:

TH is an element on the sample space for first rolling a die and then tossing a coin. A coin has both heads and tails.

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3 3/4 times 2 1/3 times 7/2 I really need some help on this

MrRa [10]

Hope this will help u...:)

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3 years ago

Read 2 more answers

If i know the area of a square, how do i find the side length?

Talja [164]

Answer:

Um im no expert but

Step-by-step explanation:

length is a type of measurement, that tells how long something is or how wide it is. hoped this helped^^

6 0

3 years ago

A bag contains 6 red marbles, 10 white marbles, and 7 blue marbles. You draw 4 marbles out at random, without replacement. What

Lapatulllka [165]

Answer:

3/1771

Step-by-step explanation:

Probability is the chance or likelihood that an event will occur.

Probability = expected outcome/total outcome

Total outcome = total number of marbles

Total outcome = 6+7+10 = 23marbles

If four marbles are drawn, the probability that they are all red is expressed as;

Pr(picking the first red marble) = 6/23

Since it is without replacement, the total marble becomes 22

Pr(picking the second) = 5/22

Pr(picking the third marbles = 4/21

Pr(picking the 4th marble) = 3/20

Note that the number of red marbles keep reducing and the total marble keep reducing as well.

Pr(picking the 4marbles) = 6/23×5/22×4/21×3/20 (we multiplied because or is an independent events)

Pr(picking the 4marbles) = 360/212520

Pr(picking 4 red marbles) = 72/42504 =36/21,252

= 9/5313

= 3/1771

Hence the probability of picking 4 marbles at random without replacement is 3/1771

4 0

3 years ago

Use mathematical induction to show that 4^n ≡ 3n+1 (mod 9) for all n equal to or greater than 0

cestrela7 [59]

When $n=0$ , you have

$4^0=1\equiv3(0)+1=1\mod9$

Now assume this is true for $n=k$ , i.e.

$4^k\equiv3k+1\mod9$

and under this hypothesis show that it's also true for $n=k+1$ . You have

$4^k\equiv3k+1\mod9$
$4\equiv4\mod9$
$\implies 4\times4^k\equiv4(3k+1)\mod9$
$\implies 4^{k+1}\equiv12k+4\mod9$

In other words, there exists $M$ such that

$4^{k+1}=9M+12k+4$

Rewriting, you have

$4^{k+1}=9M+9k+3k+4$
$4^{k+1}=9(M+k)+3k+3+1$
$4^{k+1}=9(M+k)+3(k+1)+1$

and this is equivalent to $3(k+1)+1$ modulo 9, as desired.

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3 years ago

How can i find out what fraction is the largest?

MariettaO [177]

By paying attention in school

4 0

4 years ago