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Zina [86]
3 years ago
5

Using the rules for the significant figures what do you get when you add 24.545 and 307.3

Physics
1 answer:
Tamiku [17]3 years ago
4 0
1
2 4. 5 4 5
+3 0 7. 3 0 0
——————
3 3 1 8 4 5
line up the decimal points and add.
hope this helps!
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MamaMia's Pizza purchases its pizza delivery boxes from a printing supplier. MamaMia's delivers on-average 200 pizzas each month
Ira Lisetskai [31]

Answer:

a) 138 units

b) 17 units

c) 17 units

d) Total Cost = $353.35

Explanation:

Given:

Average pizzas delivered = 200

Charge of inventory holding = 30% of cost

Lead time = 7 days

Now,

a) Economic Order Quantity =  \sqrt\frac{2\times\textup{Annual Demand}\times\textup{Cost per Order}}{\textup{Carrying cost}}

also,

Annual Demand = 200 × 12 = 2400

Cost per Order = Cost of Box + Processing Costs

= 30 cents + $10

= $10.30

and, Carrying Cost = \frac{\textup{Total Inventory Cost}}{\textup{total annual demand}}

=\frac{\textup{Total Cost per order}\times\textup{Annual demand}\times\frac{25}{100}}{\textup{Annual demand}}

= \frac{\$10.30\times2400}\times\frac{25}{100}}{2400}

= $2.575

Therefore,

Economic Order Quantity =  \sqrt\frac{2\times\textup{2400}\times\textup{10.30}}{\textup{2.575}}

= 138.56 ≈ 138 units

b) Reorder Point

= (average daily unit sales × the lead time in days) + safety stock

= (\frac{200}{30}\times7

= 46.67 ≈ 47 units

c) Number of orders per year = \frac{\textup{Annual Demand}}{\textup{Economic order quantity}}

= \frac{\textup{2400}}{\textup{138}}

= 17.39 ≈ 17 units

d) Total Annual Cost (Total Inventory Cost)

= Ordering Cost + Holding Cost

Now,

The ordering Cost = Cost per Order × Total Number of orders per year

= $10.30 × 17

= $175.1

and,

Holding Cost = Average Inventory Held × Carrying Cost per unit

Average Inventory Held = \frac{\textup{0+138}}{\textup{2}} =  69

Carrying Cost per unit = $2.575

Holding Cost = 69 × $2.575 =  

$177.675

Therefore,

Total Cost = Ordering Cost + carrying cost

= $175.1  + $177.675 = $353.35

5 0
3 years ago
Read 2 more answers
a football player kicks a ball with a mass of 0.42kg. The average acceleration of the football was 14.8 m/s2. How much force did
Zarrin [17]

Answer:

6.216 N

Explanation:

As for Newton's second law of motion

F=ma

where F= the acting force

           m=subjected mass

           a= the acceleration

applying F=ma to the football

F=m*a

 =0.42*14.8

 =6.216 N

6.216 N of a force is supplied to the ball

5 0
2 years ago
Batteries involve the flow of electricity from one metal to another. What happens to these metals to cause the batteries to, eve
Nataly [62]
The metals will start to rust lol. i think. because this messes up how the metals conduct the flow of the electricity.
3 0
3 years ago
What are non-clastic?
igor_vitrenko [27]
Non clastic sedimentary rocks from chemical reactions, chiefly in the ocean. Nonclastic and clastic sedimentary rocks are the only members of the rock family that contain fossils as well as indicators of the climate that was present when the rock was formed.
8 0
3 years ago
A merry-go-round with a rotational inertia of 600 kg m2 and a radius of 3.0 m is initially at rest. A 20 kg boy approaches the m
nekit [7.7K]

Answer:

The velocity of the merry-go-round after the boy hops on the merry-go-round is 1.5 m/s

Explanation:

The rotational inertia of the merry-go-round = 600 kg·m²

The radius of the merry-go-round = 3.0 m

The mass of the boy = 20 kg

The speed with which the boy approaches the merry-go-round = 5.0 m/s

F_T \cdot r = I \cdot \alpha  = m \cdot r^2  \cdot \alpha

Where;

F_T = The tangential force

I =  The rotational inertia

m = The mass

α = The angular acceleration

r = The radius of the merry-go-round

For the merry go round, we have;

I_m \cdot \alpha_m  = I_m \cdot \dfrac{v_m}{r \cdot t}

I_m = The rotational inertia of the merry-go-round

\alpha _m = The angular acceleration of the merry-go-round

v _m = The linear velocity of the merry-go-round

t = The time of motion

For the boy, we have;

I_b \cdot \alpha_b  = m_b \cdot r^2  \cdot \dfrac{v_b}{r \cdot t}

Where;

I_b = The rotational inertia of the boy

\alpha _b = The angular acceleration of the boy

v _b = The linear velocity of the boy

t = The time of motion

When the boy jumps on the merry-go-round, we have;

I_m \cdot \dfrac{v_m}{r \cdot t} = m_b \cdot r^2  \cdot \dfrac{v_b}{r \cdot t}

Which gives;

v_m = \dfrac{m_b \cdot r^2  \cdot \dfrac{v_b}{r \cdot t} \cdot r \cdot t}{I_m} = \dfrac{m_b \cdot r^2  \cdot v_b}{I_m}

From which we have;

v_m =  \dfrac{20 \times 3^2  \times 5}{600} =  1.5

The velocity of the merry-go-round, v_m, after the boy hops on the merry-go-round = 1.5 m/s.

5 0
2 years ago
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