The rare tweety-birds are decreasing at a rate proportional to the

One number is larger than another by 9. If the greater number is increased by 10, and the lesser number is tripled, the sum of t

kkurt [141]

Answer:

The larger number is -6, the smaller number is -15

Step-by-step explanation:

We have two numbers, a and b.

We know that one number is larger than another by 9.

Then we can write:

a = b + 9

then a is larger than b by 9 units.

If the greater number is increased by 10 (a + 10) and the lesser number is tripled (3*b), the sum of the two would be -41:

(a + 10) + 3*b = -41

So we got two equations:

a = b + 9

(a + 10) + 3*b = -41

This is a system of equations.

One way to solve this is first isolate one variable in one of the two equations:

But we can see that the variable "a" is already isolated in the first equation, so we have:

a = b + 9

now we can replace that in the other equation:

(a + 10) + 3*b = -41

(b + 9) + 10 + 3*b = -41

now we can solve this for b.

9 + b + 10 + 3b = -41

(9 + 10) + (3b + b) = -41

19 + 4b = -41

4b = -41 -19 = -60

b = -60/4 = -15

b = -15

then:

a = b + 9

a = -15 + 9 = -6

a = -6

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

Any one that knows this please help thanks!

OLga [1]

Replace each occurrence of x by 7. So we have:-

7^2 + 3 / 7 - 9

= 52 / -2

= -26 answer

5 0

2 years ago

Join my zoom anyone codess one the screeen join

Vesna [10]

Answer:

I can but why?

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers

Six times a number a number increased by three is negative twenty-one. Find the number and write the equation

salantis [7]

let number be x

6x+3= -21

6x=-24

x=-4

3 0

2 years ago