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3 years ago

What is the HCF of 7 and 13

1 answer:

3 0

Answer:

The HCF of 7 and 13 is 1

Step-by-step explanation:

Prime numbers only have one factor which is 1 and itself
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Which of following sets is equivalent to (a,b,c,d,e)

borishaifa [10]

Answer:

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Step-by-step:

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2 years ago

there are 5 people at a party. each person must shake hands once with every other person at the party. how many handshakes does

Viefleur [7K]

Here's why:

Say the 5 people are A, B, C, D, E 

A shakes hands w/ B C D E 
B shakes w/ C D E 
C with D E 
D with E 

That's 10

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3 years ago

2x=3y+1/2 I have to write this in standard form and slope intercept form

Sav [38]

Answer:

Step-by-step explanation:

2x = 3y + 1/2

standard : Ax + By = C

2x = 3y + 1/2....multiply everything by 2 to get rid of the fractions

4x = 6y + 1 ....subtract 6y from both sides

4x - 6y = 1 <==

slope intercept : y = mx + b

2x = 3y + 1/2....subtract 1/2 from both sides

2x - 1/2 = 3y....divide everything by 3

2/3x - 1/6 = y...rearrange

y = 2/3x - 1/6 <===

8 0

3 years ago

A minor league baseball team plays 128 games a seanson. If the tam won 16 more than three times as many games as they lost how m

tiny-mole [99]

Answer: The team won 100 games and losses 28 games

Step-by-step explanation:

Let the number of games won denoted by x and the games lost is denoted by y .

Given : A minor league baseball team plays 128 games in a season.

Team won 16 more than three times as many games as they lost.

Then, the system of equations, we have

$x+y=128-------(1)\\\\x=3y+16------------(2)$

We can rewrite equation (2) as $x-3y=16----------------(3)$

Now, Eliminate equation (3) from (1), we get

$y-(-3y)=128-16\\\\\Rightarrow\ 4y=112\\\\\Rightarrow\ y=\dfrac{112}{4}=28$

Substitute the value of y in (2), we get

$x=3(28)+16=84+16=100$

Hence, the team won 100 games and losses 28 games.

7 0

3 years ago

For the month of March in a certain city, 57% of the days are cloudy. Also in the month of March in the same city, 55% of the da

oksian1 [2.3K]

Answer: 0.9649

Step-by-step explanation:

Let A denote the event that the days are cloudy and B denotes the event that the days are rainy.

Given : For the month of March in a certain city, the probability that days are cloudy : $P(A)=0.57$

Also in the month of March in the same city,, the probability that the days are cloudy and rainy : $P(A\cap B)=0.55$

Now by using the conditional probability, the probability that a randomly selected day in March will be rainy if it is cloudy will be :-

$P(B|A)=\dfrac{P(A\cap B)}{P(A)}$

$\Rightarrow\ P(B|A)=\dfrac{0.55}{0.57}\\\\=0.964912280702\approx0.9649\ \ \text{[Rounded to four decimal places.]}$

Hence, the probability that a randomly selected day in March will be rainy if it is cloudy = 0.9649

4 0

3 years ago

What is the HCF of 7 and 13​

What is the HCF of 7 and 13