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3 years ago

What is the prime factorization of 120?

Mathematics

2 answers:

aleksklad [387]3 years ago

8 0

A is the answer hdndnbdbfb

Sergio039 [100]3 years ago

4 0

he prime factorization of 120 is 2*2*2*3*5 .

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The difference between 48 and 13 algebra numeral expression.

Sphinxa [80]

Numerical expressions contain numbers, while algebraic expressions contain variables and numbers.

Numerical Expression

"Difference" indicates that we'll be subtracting 13 from 48.

48 - 13 = 35

Algebraic Expression

Variables represent the unknown number, in this case the difference between 48 and 13. Let d represent the difference between the two.

48 - 13 = d

35 = d

6 0

3 years ago

The friendly sausage factory (fsf) can produce hot dogs at a rate of 5,000 per day. fsf supplies hot dogs to local restaurants a

juin [17]

Answer:

a. 21 327 hot dogs/run

b. 70 runs/yr

c. 4 da/run

Step-by-step explanation:

Data:

Production rate (p) = 5000/da

Usage rate (u) = 260/da

Setup cost (S) = $66

Annual carrying cost (H) = $0.45/hot dog

Production days (d) = 294 da

Calculations:

a. Optimal run size

(i) Annual demand (D) = pd = (5000 hot dogs/1 day) × (294 days/1 yr)

= 1 470 000 hot dogs/yr

(ii) Economic run size

$Q_{0}= \sqrt{\frac{2DS }{ h}\times\frac{ p}{p-u }}$

$= \sqrt{\frac{2\times1470000\times66 }{ 0.45}\times\frac{ 5000}{5000-260 }}$

$= \sqrt{431200000\times\frac{ 5000}{4740 }}$

$= \sqrt{454852321}$

= 21 327 hot dogs/run

b. Number of runs per year

Runs = D/Q₀ = (1 470 000 hot dogs/1yr) × (1 run/21 327 hotdogs)

= 70 runs/yr

c. Length of a run

Length = Q₀/p = (21 327 hot dogs/1 run) × (1 da/5000 hot dogs)

= 4 da/run

8 0

3 years ago

Probability that he will need to purchase at least three cups

alukav5142 [94]

Need more info for this question

5 0

4 years ago

A 200-liter tank initially full of water develops a leak at the bottom. Given that 20% of the water leaks out in the first 5 min

Leya [2.2K]

Answer:

127.53 liters left after 10 minutes

Step-by-step explanation:

Let

$A \to Amount$

$t \to time$

Given

$A(0) = 200$ --- initial

$A(5) = 200 * (1 - 20\%) = 160$ --- the amount left, after 5 minutes

Required

$A(10)$ --- amount left after 5 minutes

To do this, we make use of:

$A(t) = A(0) * e^{kt}$

$A(5) = 160$ implies that:

$160 = 200 * e^{k*5}$

Divide both sides by 200

$0.80 = e^{k*5}$

Take natural logarithm of both sides

$\ln(0.80) = \ln(e^{k*5})$

$\ln(0.80) = \ln(e^{5k})$

$\ln(0.80) = 5k\ln(e)$

So, we have:

$-0.223 = 5k$

Divide by 5

$k = -0.045$

So, the function is:

$A(t) = A(0) * e^{kt}$

$A(t) = 200 * e^{-0.045t}$

The amount after 10 minutes is:

$A(10) = 200 * e^{-0.045*10}$

$A(10) = 200 * e^{-0.45}$

$A(10) = 127.53$

3 0

3 years ago

The points (6, - 5) and (4,p) fall on a line with a slope of - 3. what is the value of p?

Hatshy [7]

The value of p would be -4.33 or -4 and 2/3.

4 0

4 years ago