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2 years ago

IM SO CONFUSED!!! I need help what is the difference between 1+0 and 1x1 HURRY DUE TODAY

Mathematics

2 answers:

Klio2033 [76]2 years ago

8 0

Answer:

one is addition the other is multiplication, though, both the sum and product equal the same solution

Step-by-step explanation:

hope it helps

Tcecarenko [31]2 years ago

3 0

Answer:

The answer is 0

Step-by-step explanation:

1+0=1

1x1=1

1-1=0

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Given: Lines, and I are parallel, with transversal line t.

VladimirAG [237]

Answer:

5 = e=h

3 = a=e

6 = b=d

4 = d=f

2 = a=h

1 = c=f

Step-by-step explanation:

8 0

2 years ago

"A cylindrical canister contains 3 tennis balls. Its height is 8.75inches, and its radius is 1.5. The diameter of one tennis bal

puteri [66]

----------------------------------------------
Find Volume of 1 tennis ball:
----------------------------------------------
Volume of 1 tennis ball = 4/3 x 3.14 x (2.5 ÷ 2)³

Volume of 1 tennis ball = 8.18 in³

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Find Volume of 3 tennis balls:
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Volume of 3 tennis balls = 8.18 x 3 = 24.54 in³

----------------------------------------------
Find Volume of teh cylindrical canister:
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Volume of the cylindrical canister = 3.14 x 1.5² x (2.5 x 3)

Volume of the cylindrical canister = 52.99 in³

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Find unoccupied space:
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Unoccupied space = 52.99 - 24.54

Unoccupied space = 28.45 in³

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Answer: 28.45 in³
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8 0

3 years ago

Read 2 more answers

Resistors are labeled 100 Ω. In fact, the actual resistances are uniformly distributed on the interval (95, 103). Find the mean

Zinaida [17]

Answer:

$E[R]$ = 99 Ω

$\sigma_R$ = 2.3094 Ω

P(98<R<102) = 0.5696

Step-by-step explanation:

The mean resistance is the average of edge values of interval.

Hence,

The mean resistance, $E[R] = \frac{a+b}{2} = \frac{95+103}{2} = \frac{198}{2}$ = 99 Ω

To find the standard deviation of resistance, we need to find variance first.

$V(R) = \frac{(b-a)^2}{12} =\frac{(103-95)^2}{12} = 5.333$

Hence,

The standard deviation of resistance, $\sigma_R = \sqrt{V(R)} = \sqrt5.333$ = 2.3094 Ω

To calculate the probability that resistance is between 98 Ω and 102 Ω, we need to find Normal Distributions.

$z_1 = \frac{102-99}{2.3094} = 1.299$

$z_2 = \frac{98-99}{2.3094} = -0.433$

From the Z-table, P(98<R<102) = 0.9032 - 0.3336 = 0.5696

5 0

3 years ago

PLEASE HELP ME! g +16/4 - 16 Simplify the variable expression by evaluating its numerical part, and enter your answer below.

Jobisdone [24]

Apex? I just did got this question and I put
g - 4
and got it right ✌️

3 0

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