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2 years ago

12

Chapter 7 of the Jiuzhang suanshu presents a problem of two linear equations involving acres of land and their respective prices

. One of the two equations can be translated to: 300x + 300 x plus StartFraction 500 over 7 EndFraction left-parenthesis 87.5 right-parenthesis equals 10,000.y = 10000 If y = 87.5, what is the value for x? 300x + 300 x plus StartFraction 500 over 7 EndFraction y equals 10,000.y = 10,000 300x + 300 x plus StartFraction 500 Over 7 EndFraction y equals 10,000.(87.5) = 10,000 300x + 6,250 = 10,000 x =

2 answers:

Nutka1998 [239]2 years ago

7 0

Answer:

12.5

Step-by-step explanation:

i just did it

iVinArrow [24]2 years ago

6 0

Answer: 12.5

Step-by-step explanation:

You might be interested in

An aircraft carrier can travel 450 km in the same time a container ship travels 300 km. If the aircraft carrier was travelling 5

Leto [7]

Answer:

10 km/hr

Step-by-step explanation:

The aircraft carrier and the container ship both took 30 hours to travel their respective distances. If you divide both numbers by 30, you find that their differences is 5. We can tell that the aircraft carrier traveled 450 ÷ 30 = 15 km/hr and the container ship traveled at a speed of 300 ÷ 30 = 10 km/hr. The aircraft carrier traveled 5 km/hr faster. So, we can tell the container ship traveled at 10 km/hr.

Hope this helps! Plz give me brainliest, it will help me achieve my next rank.

6 0

2 years ago

Type the correct answer in each box. Use numerals instead of words. If necessary, use / for the fraction bar(s).

Tanzania [10]

Answer:

Approximately, 159 men weighs more than 165 pounds and 159 men weighs less than 135 pounds.

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 150 pounds

Standard Deviation, σ = 15

We are given that the distribution of weights of 1000 men is a bell shaped distribution that is a normal distribution.

Formula:

$z_{score} = \displaystyle\frac{x-\mu}{\sigma}$

P( men weighing more than 165 pounds)

P(x > 165)

$P( x > 165) = P( z > \displaystyle\frac{165 - 150}{15}) = P(z > 1)$

$= 1 - P(z \leq 1)$

Calculation the value from standard normal z table, we have,

$P(x > 165) = 1 - 0.8413 = 0.1587 = 15.87\%$

Approximately, 159 men weighs more than 165 pounds.

P(men weighing less than 135 pounds)

P(x < 135)

$P( x < 135) = P( z < \displaystyle\frac{135 - 150}{15}) = P(z < -1)$

Calculation the value from standard normal z table, we have,

$P(x < 135) = 0.1587 = 15.87\%$

Approximately, 159 men weighs less than 135 pounds.

6 0

2 years ago

a square painting has an area of 81x^2-90x-25. A second square painting has an area of 25x^2+30x+9. What is an expression that r

galina1969 [7]

Answer:

The answer in the procedure

Step-by-step explanation:

Let

A1 ------> the area of the first square painting

A2 ----> the area of the second square painting

D -----> the difference of the areas

we have

$A1=81x^{2}-90x-25$

$A2=25x^{2}+30x+9$

case 1) The area of the second square painting is greater than the area of the first square painting

The difference of the area of the paintings is equal to subtract the area of the first square painting from the area of the second square painting

D=A2-A1

$D=(25x^{2}+30x+9)-(81x^{2}-90x-25)$

$D=(-56x^{2}+120x+34)$

case 2) The area of the first square painting is greater than the area of the second square painting

The difference of the area of the paintings is equal to subtract the area of the second square painting from the area of the first square painting

D=A1-A2

$D=(81x^{2}-90x-25)-(25x^{2}+30x+9)$

$D=(56x^{2}-120x-34)$

4 0

2 years ago

Lawrence has two part-time jobs one at the movie theater and wanted a pet store he's paid by the hour at each job last week he w

alexandr402 [8]

Let x be the earnings/hour at the movie theater.
y be the earnings/hour at the pet store

Systems of equations:

$152.50 = 10x + 8y
$171 = 8x + 10y

Solving for x and y:
x = $4.36/ hour in the movie theater
y = $13.61/ hour in the pet store.<span />

8 0

2 years ago

Suppose that demand in period 1 was 7 units and the demand in period 2 was 9 units. Assume that the forecast for period 1 was fo

Zepler [3.9K]

Answer:

Step-by-step explanation:

Forecast for period 1 is 5

Demand For Period 1 is 7

Demand for Period 2 is 9

Forecast can be given by

$F_{t+1}=F_t+\alpha (D_t-F_t)$

where

$F_{t+1}=Future Forecast$

$F_t=Present\ Period\ Forecast$

$D_t=Present\ Period\ Demand$

$\alpha =smoothing\ constant$

$F_{t+1}=5+0.2(7-5)$

$F_{t+1}=5.4$

Forecast for Period 3

$F_{t+2}=F_{t+1}+\alpha (D_{t+1}-F_{t+1})$

$F_{t+2}=5.4+0.2\cdot (9-5.4)$

$F_{t+2}=6.12$

8 0

2 years ago