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3 years ago

Please show your work:

Mathematics

1 answer:

Shalnov [3]3 years ago

6 0

Answer:

b . 4/5

60 :15 minute : 4

3 /15 x 4 = 12/15 = 4/5

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Use an area model to solve. 2 2/3 x 1 1/3?

natima [27]

<span>the answer is : 9.77777777778</span>

6 0

3 years ago

Read 2 more answers

The A&M Hobby Shop carries a line of radio-controlled model racing cars. Demand for the cars is assumed to be constant at a

Bond [772]

Answer:

Step-by-step explanation:

A) Demand per month= 40 cars

Annual Demand (D)= 12*40 = 480

Fixed Cost per order (K)= 15

Holding Cost= 20% of cost= 60 *0.2 = 12

a. Economic Order Quantity=

Q^{*}={\sqrt {{\frac {2DK}{h}}}}

= √(2*480*15)/12

=34.64 ~ 35

Total Cost =P*D+K(D/EOQ)+h(EOQ/2) P= Cost per unit

= 60*480+ 15(480/35) + 12(35/2)

= 28800+ 205.71+ 210

=$29215.71

B). Backorder Cost (b)= $45

Qbo= Q* × √( b+h/ h)

= 35*√(12+45/ 45)

= 35* 1.12

=39.28 ~ 39

Shortage (S)= Qbo * (K/K+b)

= 39* (15/15+45)

= 39* 0.25

= 9.75

Total Cost Minimum=( bS2/ 2Qbo) + P (Qbo- S)2/2Qbo + K(D/Qbo)

=45* 9.752 / 2* 392 + 60 (39-9.75)2/ 2* 392 + 15 ( 480/39)

= 1.40+ 21.9.+ 184.61

=$207.91

C)Length of backorder days (d) = Demand ÷ amount of working days

d = 480 ÷ 300

d = 1.6

Calculate the backorders as the maximum number of backorders divided by the demand per day

s/d = 9.75/1.6 = 6.09 days (answer)

D) Calculate the difference in total between not using backorder:

$207.85 + $207.85 - 207.91 = $207.79

The saving in using backorder is $207.79.

Therefore I would recommend using a backorder

6 0

2 years ago

A random sample of 28 statistics tutorials was selected from the past 5 years and the percentage of students absent from each on

SVEN [57.7K]

Answer:

Step-by-step explanation:

Hello!

X: number of absences per tutorial per student over the past 5 years(percentage)

X≈N(μ;σ²)

You have to construct a 90% to estimate the population mean of the percentage of absences per tutorial of the students over the past 5 years.

The formula for the CI is:

X[bar] ± $Z_{1-\alpha /2}$ * $\frac{S}{\sqrt{n} }$

⇒ The population standard deviation is unknown and since the distribution is approximate, I'll use the estimation of the standard deviation in place of the population parameter.

Number of Absences 13.9 16.4 12.3 13.2 8.4 4.4 10.3 8.8 4.8 10.9 15.9 9.7 4.5 11.5 5.7 10.8 9.7 8.2 10.3 12.2 10.6 16.2 15.2 1.7 11.7 11.9 10.0 12.4

X[bar]= 10.41

S= 3.71

$Z_{1-\alpha /2}= Z_{0.95}= 1.645$

[10.41±1.645* $(\frac{3.71}{\sqrt{28} } )$ ]

[9.26; 11.56]

Using a confidence level of 90% you'd expect that the interval [9.26; 11.56]% contains the value of the population mean of the percentage of absences per tutorial of the students over the past 5 years.

I hope this helps!

7 0

3 years ago

3. The student council is selling roses to collect funds for a charity. The graph be

dexar [7]

Answer:

$Unit = 3$