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2 years ago

Focus on number two ok

Mathematics

2 answers:

melomori [17]2 years ago

7 0

Its simple, convert 1 and 2/3 into a fraction which is 5/3 and use KCF which is Keep Change and Flip,

How this works is that you keep the first fraction, then change the division sign to a multiplication sign (Change = change the sign into its oppisite) then flip the other fraction from being 5/3 into 3/5. And then you multiply top times top and bottom times bottom. After that you simplify if you can, Hope this helped!

mr Goodwill [35]2 years ago

7 0

anwser is 5/12

dividing a fraction is the same as multiplying by a reciprocal

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3 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

The numerator and denominator of a fraction are in the ratio of 3 to 5. If the numerator and denominator are both increased by 2

ivolga24 [154]

Question:

The numerator and denominator of a fraction are in the ratio of 3 to 5. If the numerator and denominator are both decreased by 2, the fraction is now equal to $\frac{1}{2}$ .

If n = the numerator and d = the denominator, which of the following systems of equations could be used to solve the problem?

5n = 3d and n - 2 = 2d - 4

5n = 3d and 2n - 4 = d - 2

3n = 5d and 2n - 4 = d - 2

Answer:

5n = 3d and 2n – 4 = d – 2

Solution:

Let n be the numerator of the fraction and d be the denominator of the fraction.

Given the numerator and denominator of a fraction are in the ratio of 3 to 5.

This can be written as n : d = 3 : 5.

⇒ $\frac{n}{d}= \frac{3}{5}$ – – – – (1)

Do cross multiplication, we get

⇒ 5n = 3d

When the numerator and denominator are decreased by 2, the fraction is equal to $\frac{1}{2}$ .

⇒ $\frac{n-2}{d-2}= \frac{1}{2}$

Do cross multiplication, we get

⇒ 2(n –2)=1(d – 2)

⇒ 2n – 4 = d – 2

Hence, 5n = 3d and 2n – 4 = d – 2 can be used to solve the problem.

6 0

3 years ago

Read 2 more answers