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VMariaS [17]
3 years ago
10

Plz someone help me with questions 2 down I really need help because I still have a different homework

Mathematics
1 answer:
sergejj [24]3 years ago
8 0

Answer:

2=0.8oz or 4/5 3=0.27 4=0.13 5=0.75 or 3/4 6=

Step-by-step explanation:

For all of these problems, you need to find the UNIT RATE

To find the unit rate, divide the numerator and denominator of the given rate by the denominator of the given rate

#2) 16/20=0.8

#3) 4.86/18=0.27

#4)2.6/20=0.13

#5)$6/8 hours=0.75

#6)? Sorry can’t see full problem on my computer screen my computers broken

I hope these help! Good luck with your other homework!

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HELP PLEASE 40 POINTS The graph shows the prices of different numbers of bushels of corn at a store in the current year. The tab
Vesna [10]

Answer:

see below

Step-by-step explanation:

Part A

Since the lines goes through the point (0,0) the graph is proportional. We can find the rate of change by take the price of corn and dividing by the number of bushels

24/3 = 8 dollars/ bushel

Part B

Previous Year Number of Bushels Price of Corn (dollars)

                                                  3 21

                                                 6 42

                                                  9 63

                                                 12 84

We can find the rate of change for the previous year by using the slope formula

m = (y2-y1)/(x2-x1)

m = (84-63)/(12-9)

    =21 / 3  

    = 7

The previous year was 7 dollars per bushel

The increase was 8-7 = 1 dollar per bushel

4 0
3 years ago
Read 2 more answers
Suppose there are 4 defective batteries in a drawer with 10 batteries in it. A sample of 3 is taken at random without replacemen
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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

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Multiplying and Dividing Radical Expressions
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