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2 years ago

Find the gcf of the monomial factor 15a-25b+20

Mathematics

1 answer:

Olin [163]2 years ago

6 0

The free websoftware for math professionals does not understand the question

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Two cards are dealt at random from a standard deck of 52 cards. What is the probability that the first card is a King and the se

34kurt

Answer:

1 / 51

Step-by-step explanation:

Given that :

Number of cards in a deck = 52

Number of heart suit = 13

Number of kings = 4

Recall:

Probability = (required outcome / Total possible outcomes)

P(first card is king) = 4/52 = 1/13

P(second card is heart) = 13/51

Hence,

P(first card is king) * P(second card is heart)

(1/ 13) * (13/51) = 13 / 663 = 1/ 51

5 0

3 years ago

The table below shows all outcomes for rolling 2 dice. The bold numbers 1 through 6 show the possible results for each die. All

Leokris [45]

Part I:

1. If you have to circle each outcome in the table that at least 1 die is a 4, then you have to circle row corresponding to the number 4 and column corresponding to the number 4.

2. If you have to cross out each outcome where the sum of the dice is greater than 5, then you have to cross out all not bold numbers 6, 7, 8, 9, 10, 11 and 12 in the table.

Part II:

Here you can see 36 possible outcomes.

1. The probability that at least 1 die is a 4 is:

$Pr(A)=\dfrac{11}{36}$ - (favorable outcomes are 4+1=5, 4+2=6, 4+3=7, 4+4=8, 4+5=9, 4+6=10, 1+4=5, 2+4=6, 3+4=7, 5+4=9, 6+4=10).

2. The probability that the sum of the dice is greater than 5 is:

$Pr(B)=\dfrac{26}{36}=\dfrac{13}{18}.$

3. The probability that at least 1 die is a 4 and that the sum of the dice is greater than 5 is

$Pr(A\cap B)=\dfrac{9}{36}=\dfrac{1}{4}.$

4. The probability that at least 1 die is a 4 or that the sum of the dice is greater than 5 is:

$Pr(A\cup B)=\dfrac{28}{36}=\dfrac{7}{9}.$

4 0

3 years ago

The length of a rectangle exceeds the breadth by 8 cm. Find the area of the rectangle, if its perimeter is 180 cm

sergiy2304 [10]

Answer:

2009cm^2

Step-by-step explanation:

Given data

let the length be L

and the width be w

w= x

L=x+8

P= 180cm

P= 2L+2W

180= 2(x+8)+2(x)

180= 2x+16+2x

collect like terms

180= 4x+16

180-16= 4x

164= 4x

x= 164/4

x=41

W= 41cm

L= 41+8

L= 49cm

A= L*W

A=49*41

A=2009cm^2

Hence the area is 2009cm^2

3 0

2 years ago

Convert each equation to standard form y-5=-3(x+2)

ale4655 [162]

You said each equation but only posted one.

The standard form takes the place of

Ax + By = C.

y-5 = -3(x+2)

y - 5 = -3x - 6

y = -3x - 6 + 5

3x + y = -1

Done.

8 0

2 years ago

Read 2 more answers

The length of a rectangle is 8 more than the width. The area is 105 square centimeters. Find the length and width of the rectang

sashaice [31]

Answer:

Width = 7 cm

Length = 15 cm

Step-by-step explanation:

Let width = w

Length = w + 8

Area of rectangle = 105 sq. cm

length * breadth = 105

(w + 8)*w = 105

Use distributive property

w² + 8w = 105

w² + 8w - 105 = 0

Sum = 8

Product = -105

Factors = 15 , -7 {15 +(-7) = 8 & 15*(-7) = -105}

w² - 7m + 15m - 105 = 0

w(w - 7) + 15(m - 7) = 0

(w - 7)(w + 15) = 0

Ignore w + 15= 0 as measurements will not be in negative value

w - 7 = 0

w = 7

Width = 7 cm

Length = 7 + 8 = 15 cm

8 0

2 years ago