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Studentka2010 [4]

3 years ago

5

Which of the following describe point c?

1 answer:

ElenaW [278]3 years ago

4 0

The answer is (-2,-1)

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Subtract x^2-6x+4 from 2x^2+4x-8

Rasek [7]

2X^2+4X-8-X^2+6X-4
X^2+4X-8+6X-4
X^2+10X-12

3 0

2 years ago

Can I pls get help on this question

ale4655 [162]

Answer:

Yes

Step-by-step explanation:

I had this exact question.

4 0

3 years ago

Amy is pulling a wagon with a force of 30 pounds up a hill at an angle of 25°. Give the force exerted on the wagon as a vector a

Fofino [41]

Answer:

Vector (ordered pair - rectangular form)

$\vec F = (27.189,12.679)\,[lbf]$

Vector (ordered pair - polar form)

$\vec F = (30\,lbf, 25^{\circ})$

Sum of vectorial components (linear combination)

$\vec F = 27.189\cdot \hat{i} + 12.679\cdot \hat{j}\,[N]$

Step-by-step explanation:

From statement we know that force exerted on the wagon has a magnitude of 30 pounds-force and an angle of 25° above the horizontal, which corresponds to the +x semiaxis, whereas the vertical is represented by the +y semiaxis.

The force ( $\vec F$ ), in pounds-force, can be modelled in two forms:

Vector (ordered pair - rectangular form)

$\vec F = \left(\|\vec F\|\cdot \cos \theta, \|\vec F\|\cdot \sin \theta\right)$ (1)

Vector (ordered pair - polar form)

$\vec F = \left(\|\vec F\|, \theta\right)$

Sum of vectorial components (linear combination)

$\vec {F} = \left(\|\vec F\|\cdot \cos \theta\right)\cdot \hat{i} + \left(\|\vec F\|\cdot \sin \theta \right)\cdot \hat{j}$ (2)

Where:

$\|\vec F\|$ - Norm of the vector force, in newtons.

$\theta$ - Direction of the vector force with regard to the horizontal, in sexagesimal degrees.

$\hat{i}$ , $\hat{j}$ - Orthogonal axes, no unit.

If we know that $\|\vec F\| = 30\,lbf$ and $\theta = 25^{\circ}$ , then the force exerted on the wagon is:

Vector (ordered pair - rectangular form)

$\vec F= \left(30\cdot \cos 25^{\circ}, 30\cdot \sin 25^{\circ}\right)\,[lbf]$

$\vec F = (27.189,12.679)\,[lbf]$

Vector (ordered pair - polar form)

$\vec F = (30\,lbf, 25^{\circ})$

Sum of vectorial components (linear combination)

$\vec F = (30\cdot \cos 25^{\circ})\cdot \hat{i} + (30\cdot \sin 25^{\circ})\cdot \hat{j}\,[N]$

$\vec F = 27.189\cdot \hat{i} + 12.679\cdot \hat{j}\,[N]$

8 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

3 years ago

A rectangle has a perimeter of 48 cm and a width of 6 cm. Use the formula P=2ℓ+2w to find the length.

Vinil7 [7]

Answer:the length is equal to 18. 2l+2w= 12+?=48. 48-12=36, 36/2=18

Step-by-step explanation:

5 0

3 years ago

Read 2 more answers