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Ilia_Sergeevich [38]
3 years ago
8

Can you plese help me with it

Mathematics
2 answers:
il63 [147K]3 years ago
5 0
Of the sixteen possibilities only one is HHHT, so the probability is 

\dfrac{1}{16}

andrey2020 [161]3 years ago
5 0
Hey there!

The answer should be 1/16
You might be interested in
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
A mixture of compounds X and Y in a 0.100-cm cell had an absorbance of 0.215 at 272 nm and 0.191 at 327 nm. Find [X] and [Y] in
Oksi-84 [34.3K]

Answer: The concentration of X is 7.13582\times 10^{-6} and the concentration of Y is 2.53159\times 10^{-5}.

Step-by-step explanation:

Since we have given that

At 272 nm, absorbance = 0.215

At 327 nm, absorbance = 0.191

As we have given that

                               Compound X             Compound Y

272                               16400                            3870

327                                3990                             6420

So, our equations becomes

16400C_1+3870c_2=0.215\\\\3990C_1+6420C_2=0.191

By solving these two equations, we get that

C_1=7.13582\times 10^{-6}\\\\C_2=2.53159\times 10^{-5}

Hence, the concentration of X is 7.13582\times 10^{-6} and the concentration of Y is 2.53159\times 10^{-5}.

5 0
3 years ago
If four times a number plus 3 is 11, what is the number? a. 4 bm 2 c. 16 d. 5​
goblinko [34]

Answer:

B. 2 (Second Option)

Step-by-step explanation:

Four times: ⇒ 4

Number:⇒  y

4y+3=11

<em>Subtract by 3 both sides of equation.</em>

<em>4y+3-3=11-3</em>

<em>Simplify.</em>

<em>11-3=8</em>

<em>4y=8</em>

<em>Divide by 4 both sides of equation.</em>

<em>4y/4=8/4</em>

<em>SImplify, to find the answer.</em>

<em>8/4=2</em>

<em>y=2 is the correct answer.</em>

<em>2 is the correct answer.</em>

6 0
3 years ago
Read 2 more answers
The graph of a function with a y-intercept of -4 and a rate of change of 10
jekas [21]
Using y=Mx+b, the answer is y=10x-4.
6 0
3 years ago
please help me if you can if your going to say I'm cheating i don't care cause i know I'm not thanks!!​
Dovator [93]
Big fat badussy balls
4 0
2 years ago
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