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3 years ago

15

Ive been stuck on this expression for a while now

1 answer:

natka813 [3]3 years ago

4 0

Answer:

xy^(2)z^(6)

Step-by-step explanation:

i hope that helps you :)

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Ginny divided the cake into 12 pieces. She gave 4 pieces to Sammy. What fraction of the cake did she have left?

In-s [12.5K]

Answer: 2/3 or 1/3

Step-by-step explanation: If we have a cake that has a total of 12 pieces, and we gave four to someone else we would have 8/12 pieces left. Now if we simplify then we would have 2/3 left.

Alternate explanation: Now if we just want to divide 4/12 then we would have 1/3.

I'm happy to help let me know if I'm wrong!

5 0

2 years ago

Solve the system:<br> 4x – 3y = -21<br> X=-y-14

Liula [17]

Answer:

The solution for the system of equations is

x=2,y=−2

Explanation:

4x−3y=14.....equation (1)

y=−3x+4.....equation (2)

Solving by substitution

Substituting equation 2 in equation 1

4x−3⋅(−3x+4)=14

4x+9x−12=14

13x=14+12

13x=26

x=2

Finding y by substituting x in equation 1

4x−3y=14

4⋅2−3y=14

8−3y=14

8−14=3y

3y=−6

y=−2

4 0

3 years ago

Read 2 more answers

2. What is the vertex of the following parabola? *

sweet [91]

Answer:

42

Step-by-step explanation:

The answer to life the universe and everything is 42

7 0

2 years ago

Suppose there are 4 defective batteries in a drawer with 10 batteries in it. A sample of 3 is taken at random without replacemen

SSSSS [86.1K]

Answer:

a.) 0.5

b.) 0.66

c.) 0.83

Step-by-step explanation:

As given,

Total Number of Batteries in the drawer = 10

Total Number of defective Batteries in the drawer = 4

⇒Total Number of non - defective Batteries in the drawer = 10 - 4 = 6

Now,

As, a sample of 3 is taken at random without replacement.

a.)

Getting exactly one defective battery means -

1 - from defective battery

2 - from non-defective battery

So,

Getting exactly 1 defective battery = ⁴C₁ × ⁶C₂ = $\frac{4!}{1! (4 - 1 )!}$ × $\frac{6!}{2! (6 - 2 )!}$

= $\frac{4!}{(3)!}$ × $\frac{6!}{2! (4)!}$

= $\frac{4.3!}{(3)!}$ × $\frac{6.5.4!}{2! (4)!}$

= $4$ × $\frac{6.5}{2.1! }$

= $4$ × $15$ = 60

Total Number of possibility = ¹⁰C₃ = $\frac{10!}{3! (10-3)!}$

= $\frac{10!}{3! (7)!}$

= $\frac{10.9.8.7!}{3! (7)!}$

= $\frac{10.9.8}{3.2.1!}$

= 120

So, probability = $\frac{60}{120} = \frac{1}{2} = 0.5$

b.)

at most one defective battery :

⇒either the defective battery is 1 or 0

If the defective battery is 1 , then 2 non defective

Possibility = ⁴C₁ × ⁶C₂ = 60

If the defective battery is 0 , then 3 non defective

Possibility = ⁴C₀ × ⁶C₃

= $\frac{4!}{0! (4 - 0)!}$ × $\frac{6!}{3! (6 - 3)!}$

= $\frac{4!}{(4)!}$ × $\frac{6!}{3! (3)!}$

= 1 × $\frac{6.5.4.3!}{3.2.1! (3)!}$

= 1× $\frac{6.5.4}{3.2.1! }$

= 1 × 20 = 20

getting at most 1 defective battery = 60 + 20 = 80

Probability = $\frac{80}{120} = \frac{8}{12} = 0.66$

c.)

at least one defective battery :

⇒either the defective battery is 1 or 2 or 3

If the defective battery is 1 , then 2 non defective

Possibility = ⁴C₁ × ⁶C₂ = 60

If the defective battery is 2 , then 1 non defective

Possibility = ⁴C₂ × ⁶C₁

= $\frac{4!}{2! (4 - 2)!}$ × $\frac{6!}{1! (6 - 1)!}$

= $\frac{4!}{2! (2)!}$ × $\frac{6!}{1! (5)!}$

= $\frac{4.3.2!}{2! (2)!}$ × $\frac{6.5!}{1! (5)!}$

= $\frac{4.3}{2.1!}$ × $\frac{6}{1}$

= 6 × 6 = 36

If the defective battery is 3 , then 0 non defective

Possibility = ⁴C₃ × ⁶C₀

= $\frac{4!}{3! (4 - 3)!}$ × $\frac{6!}{0! (6 - 0)!}$

= $\frac{4!}{3! (1)!}$ × $\frac{6!}{(6)!}$

= $\frac{4.3!}{3!}$ × 1

= 4×1 = 4

getting at most 1 defective battery = 60 + 36 + 4 = 100

Probability = $\frac{100}{120} = \frac{10}{12} = 0.83$

3 0

2 years ago

What are the first 6 nonzero multiples of 15?

Sphinxa [80]

<span>5,10,15,20,25,30,35,40,45,50,55,60 or 6,12,18,24,30,36,42,48,54,60,66,72</span>

8 0

3 years ago

Read 2 more answers