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lora16 [44]
2 years ago
12

Please help i dont understand

Mathematics
1 answer:
son4ous [18]2 years ago
5 0

Answer:

\frac{ {y}^{2} }{4 {x}^{5} }

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At wendys, 3 baconators and 2 frosty cost $21.50. If 5 baconators and 1 frosty cost 30$, find the price of 1 baconnator and 1 fr
Gnom [1K]

Answer:

are you at wendys

Step-by-step explanation:

30/5=6 so 1 frosy is 6 dollars and 30-6=24 and 24/5=4.80

7 0
2 years ago
Read 2 more answers
How do you solve -k/5 times 7/5
Firdavs [7]

Answer:

\frac{-7}{25}k

Step-by-step explanation:

\frac{-k}{5} *\frac{7}{5}

just multiply across

-k*7=-7k

5*5=25

\frac{-7}{25}k

hope this helps

5 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 is the equation of the line that is parallel to the given
Orlov [11]

Answer:

x=2,-8

Step-by-step explanation:

4 0
3 years ago
What adds to 17 but multiplies to -200
ollegr [7]
-8 + 25= 17
-8 • 25 =-200
4 0
3 years ago
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