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3 years ago

11

X When I divide my number by 3 and subtract four, I get six. What is my number?

2 answers:

Neko [114]3 years ago

7 0

Answer:

your number is 30

Step-by-step explanation:

x ÷ 3 - 4 = 6

x = 30

Fittoniya [83]3 years ago

4 0

Answer:

30

is the answer

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Solve each equation and match the solution

8_murik_8 [283]

Answer:

a

Step-by-step explanation:

sub to yfbg jay

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3 years ago

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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

3 years ago

A local little league has a total of 60 players, of whom 20% are right-.handed. How many right-handed players are there?

allochka39001 [22]

Answer:

20% of 60 is 12. 12 of the players are right-handed.

7 0

3 years ago

What is the solution to the system of linear equations graphed below?

olganol [36]

Answer: $(3\frac{1}{2}, -4)$

Step-by-step explanation:

The solution to the graph of a system of equations is where they intercept. Here, it is at about $(3\frac{1}{2}, -4)$ .

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3 years ago

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What is the range of the function f(x) = 3x2 + 6x – 8?

timama [110]

Answer:

Range is {y | y ≥ –11}

Step-by-step explanation:

This is quadratic equation.

<em>A quadratic equation's range can be found if we find the vertex.</em>

For quadratic equations that have a positive number in front of $x^{2}$ , it is upward opening and thus <u>all the numbers greater than or equal to the minimum value of vertex is the range.</u>

The formula for vertex of a parabola is:

Vertex = $(-\frac{b}{2a}, f(-\frac{b}{2a})$

Where,

$a$ is the coefficient of $x^{2}$
$b$ is the coefficient of $x$

From our equation given, $a=3$ and $b=6$

Now, $x$ coordinate of vertex is $x=-\frac{b}{2a}\\x=-\frac{6}{(2)(3)}\\x=-\frac{6}{6}\\x=-1$

$y$ coordinate of the vertex IS THE MINIMUM VALUE that we want. We get this by plugging in the $x$ value [ $x=-1$ ] into the equation. So we have:

$3(-1)^{2}+6(-1)-8\\=-11$

Hence, the range would be all numbers greater than or equal to $-11$

Third answer choice is the right one.

6 0

3 years ago

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