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4 years ago

5

What is the distance between ( 2 1/2, 3) and (1, -3)

1 answer:

belka [17]4 years ago

4 0

Answer:

11.2361

explanation is up

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A storage tank is made of a 4 m tall cylinder joined by a 3 m tall cone . If the diameter of the cylinder is 5 m, what is the ra

egoroff_w [7]

Answer: The answer is 0.625 (the height of the cylinder is not needed!)

8 0

3 years ago

Simplify the expression 32a769. Assume a = 0 and b +0.<br> Enter the correct answer in the box.

Solnce55 [7]

Answer:

2a^2b^3 ^3squareroot4a

Step-by-step explanation:

4 0

3 years ago

A uniform 41.0 kg scaffold of length 6.6 m is supported by two light cables, as shown below. A 74.0 kg painter stands 1.0 m from

Nimfa-mama [501]

Step-by-step explanation:

Let

$m_p$ = mass of the painter

$m_s$ = mass of the scaffold

$m_e$ = mass of the equipment

$T$ = tension in the cables

In order for this scaffold to remain in equilibrium, the net force and torque on it must be zero. The net force acting on the scaffold can be written as

$3T = (m_p + m_s + m_e)g\:\:\:\:\:\:\:(1)$

Set this aside and let's look at the net torque on the scaffold. Assume the counterclockwise direction to be the positive direction for the rotation. The pivot point is chosen so that one of the unknown quantities is eliminated. Let's choose our pivot point to be the location of $m_e$ . The net torque on the scaffold is then

$T(1.4\:\text{m}) + m_sg(1.9\:\text{m}) + m_pg(4.2\:\text{m}) - 2T(5.2\:\text{m}) = 0$

Solving for T,

$9T = m_sg(1.9\:\text{m}) + m_pg(4.2\:\text{m})$

or

$T = \frac{1}{9}[m_sg(1.9\:\text{m}) + m_pg(4.2\:\text{m})]$

$\:\:\:\:= 423.3\:\text{N}$

To solve for the the mass of the equipment $m_e$ , use the value for T into Eqn(1):

$m_e = \dfrac{3T - (m_p + m_s)g}{g} = 14.6\:\text{kg}$

6 0

3 years ago

(2 5/8 + 5/6 + 1 1/8

VLD [36.1K]

<h2> $2\dfrac{5}{8}+\dfrac{5}{6}+1\dfrac{1}{8}$ = $3\dfrac{19}{12}$ or, $\dfrac{55}{12}$ </h2>

Step-by-step explanation:

We have,

$2\dfrac{5}{8}+\dfrac{5}{6}+1\dfrac{1}{8}$

To find, the value of $2\dfrac{5}{8}+\dfrac{5}{6}+1\dfrac{1}{8}$ = ?

∴ $2\dfrac{5}{8}+\dfrac{5}{6}+1\dfrac{1}{8}$

= $2+\dfrac{5}{8}+\dfrac{5}{6}+1+\dfrac{1}{8}$

= $(2+1)+(\dfrac{5}{8}+\dfrac{5}{6}+\dfrac{1}{8})$

= 3 + $\dfrac{15+20+3}{24}$

= 3 + $\dfrac{19}{12}$

= $3\dfrac{19}{12}$ or, $\dfrac{55}{12}$

∴ $2\dfrac{5}{8}+\dfrac{5}{6}+1\dfrac{1}{8}$ = $3\dfrac{19}{12}$ or, $\dfrac{55}{12}$

6 0

3 years ago

With one method of a procedure called acceptance sampling, a sample of items is randomly selected without replacement and the e

Veseljchak [2.6K]

Answer: Probability that the entire batch will be accepted is 0.658.

Yes, this suggests entire batch of good CDs as it is the only condition to get accepted.

Step-by-step explanation:

Since we have given that

Number of CDs = 507

Number of defective = 65

Since the entire batch will be accepted.

So, the probability of success is p = $\dfrac{507-65}{507}=\dfrac{442}{507}=0.87$

and the probability of failure is q = $\dfrac{65}{507}=0.128$

Number of CDs selected = 3

So, Probability that the entire batch will be accepted is given by

$P(X=3)=^3C_3(0.87)^3\\\\P(X=3)=0.658$

Hence, Probability that the entire batch will be accepted is 0.658.

Yes, this suggests entire batch of good CDs as it is the only condition to get accepted.

6 0

3 years ago