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3 years ago

Help with math please?

Mathematics

2 answers:

Rina8888 [55]3 years ago

8 0

Distribution
Addition
Subtraction
Division

elena-s [515]3 years ago

3 0

Distribution
Addition
Subtraction
Division

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ExtremeBDS [4]

It’s 13.1 ft I know cause I did the math all the way it’s 13.1 so yeah

5 0

3 years ago

Determine the solution set of each equation

maks197457 [2]

A=x1 = -8,x2 =2, B=x1= -5,x2=5, C=y1 = -4, y2 =5, D=y 18 over 5

alternative form

y=3 3over 5, =3.6

And E=x1=0, x2 =2 over 3

Alternative form

x1=0, x2=0.666667.

question A

Solving steps:Solve the quadratic equation

x²+6x-16=0

write 6x as a difference

x2 +8x-2x-16=0

factor out x for the expression

xx(x+8)-2x-16=0

factor out -2 from the expression

xx(x+8)- 2(x+8)=0

factor out x+8 from the expression

(x+8)x(x- 2)=0

when the product of factors equals 0,at least one factor is 0

x+8=0

x - 2=0

solve the equation for x

x= -8

x - 2=0

solve the equation for x

x= -8

x =2

the equation has 2 solution

x1 = -8,X2 =2

answer is

x1 = -8,x2 =2

Solving steps:Number of solution

x²+6x-16=0

Determine the number of solution using the

discriminant D =b²- 4ac

D =6²- 4x1x(-16)

simply the expression

D=100

Since D>0, the quadratic equation has 2 real solution

2 real solution

Answer is

2 real solution

Question B

Solving steps:Solve the quadratic equation

x²-25=0

Move the constant to the right-hand side and change its sign

x²=25

Take the square root of both sides of the equation and remember to use both positive and negative roots

x=+5

Write the solutions,one with a +sign and one with a -sign

x= -5

x=5

The equation has 2 solutions

x1= -5,x2=5

the answer is

x1= -5,x2=5

Solving steps:Number of solution

x²-25=0

Determine the number of solutions using the discriminant D=b²- 4ac

D=0²- 4x1x(-25)

Simply the expression

D=100

Since D>0,the quadratic equation has 2real solutions

Answer is

2real solutions

answer of question C is

y1 = -4, y2 =5

Answer of question D is

y 18 over 5

alternative form

y=3 3over 5, =3.6

Answer of question E is

x1=0, x2 =2 over 3

Alternative form

x1=0, x2=0.666667

Please mark as brainiest

3 0

3 years ago

Read 2 more answers

HELP PLEASE I NEED TO SUBMIT MY TEST!!

8090 [49]

Answer: 20

Step-by-step explanation:

7 0

3 years ago

Suppose that two openings on an appellate court bench are to be filled from current municipal court judges. The municipal court

Ksju [112]

Answer:

$(a)\dfrac{92}{117}$

$(b)\dfrac{8}{39}$

$(c)\dfrac{25}{117}$

Step-by-step explanation:

Number of Men, n(M)=24

Number of Women, n(W)=3

Total Sample, n(S)=24+3=27

Since you cannot appoint the same person twice, the probabilities are <u>without replacement.</u>

(a)Probability that both appointees are men.

$P(MM)=\dfrac{24}{27}X \dfrac{23}{26}=\dfrac{552}{702}\\=\dfrac{92}{117}$

(b)Probability that one man and one woman are appointed.

To find the probability that one man and one woman are appointed, this could happen in two ways.

A man is appointed first and a woman is appointed next.
A woman is appointed first and a man is appointed next.

P(One man and one woman are appointed) $=P(MW)+P(WM)$

$=(\dfrac{24}{27}X \dfrac{3}{26})+(\dfrac{3}{27}X \dfrac{24}{26})\\=\dfrac{72}{702}+\dfrac{72}{702}\\=\dfrac{144}{702}\\=\dfrac{8}{39}$

(c)Probability that at least one woman is appointed.

The probability that at least one woman is appointed can occur in three ways.

A man is appointed first and a woman is appointed next.
A woman is appointed first and a man is appointed next.
Two women are appointed

P(at least one woman is appointed) $=P(MW)+P(WM)+P(WW)$

$P(WW)=\dfrac{3}{27}X \dfrac{2}{26}=\dfrac{6}{702}$

In Part B, $P(MW)+P(WM)=\frac{8}{39}$

Therefore:

$P(MW)+P(WM)+P(WW)=\dfrac{8}{39}+\dfrac{6}{702}\\$P(at least one woman is appointed)=\dfrac{25}{117}$

5 0

3 years ago

The expression 10a+6c gives the cost (in dollars) for a adults and c children to eat at a buffet restaurant.

lys-0071 [83]

Yikez that’s really hard

4 0

3 years ago