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3 years ago

6

Is three a solution of z^2 +1=4+3z?

1 answer:

DiKsa [7]3 years ago

8 0

Answer:

no

Step-by-step explanation:

z^2 +1=4+3z

Substitute into the equation and see if it is true

3^2 +1 = 4+3(3)

9+1 = 4+9

10 = 13

This is not a true statement so this is not a solution

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What is this plz help me

iren [92.7K]

20 is the answer for x

8 0

2 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

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

Im having trouble figuring out the missing length and solving this, can I have some help?

belka [17]

I think the answer is 17

8 0

2 years ago

ΔEFG is located at E (0, 0), F (−7, 4), and G (0, 8). Which statement correctly classifies ΔEFG?

MariettaO [177]

Answer:

ΔEFG is an isosceles triangle.

Step-by-step explanation:

Given:

E (0, 0),

F (−7, 4),

G (0, 8)

ΔEFG

Solution:

Distance formula

Distance d = $\sqrt{(x_2-x_1)^2 +( y_2-y_1)^2$

Step 1: Finding the length of EF

By using distance formula,

$EF = \sqrt{(-7 - 0)^2 + (4-0)^2}$

$EF = \sqrt{(49) + (16)}$

$EF = \sqrt{(49) + (16)}\\EF = \sqrt{65}\\$

Step 2: Finding the length of FG

By using distance formula,

$FG = \sqrt{(0-(-7))^2+(8-4)^2}\\FG = \sqrt{(7)^2 +(4)^2}\\FG = \sqrt{49 +16}\\FG = \sqrt{65}$

Step 2: Finding the length of GE

$GE= \sqrt{(0-0)^2 + (0-8)^2}\\\\GE =\sqrt{(-8)^2}\\GE = \sqrt{64}\\GE = 8$

Thus we could see that the sides EF = FG

So it is a isosceles triangle.

6 0

3 years ago

Un número entero de tres dígitos es igual a 25 veces la suma de sus dígitos. Si los tres dígitos se invierten, el número que res

Pavlova-9 [17]

La repuesta es 375，porque 3+7+5 es 15，y 15x25 es 375. Tambien el digito de las decenas es uno menos que la suma de los digitos de las centenas y de las unidades.

5 0

3 years ago