Help please Algebra 2 B

A carpenter needs to cut four sections, each 3–feet 8–inches long, from a piece of molding. If the board is only sold by the foo

Volgvan

3 feet 8 inches converted to inches is 3x12+8=44inches

44inches times 4 pieces is 176 inches.

176 inches converted back to feet and inches is 176/12=14 feet 8 inches so he would need 15 feet of board

3 0

3 years ago

6.1.8 The following set of scores was obtained from a quiz: 4, 5, 8, 9, 11, 13, 15, 18, 18, 18, 20. The teacher computes the usu

Inessa05 [86]

Answer:

B. Median

D. IQ R

Step-by-step explanation:

from the given data :

mean = 12.6364 , median =13,

st. dev = 5.6262

Q3 = 18 , Q1 = 8

IQ R = Q3 - Q1 = 18-8 = 10

One of the 18s should be 16 then,

mean = 12.4545, median = 13,

st. dev = 5.4656

Q3 = 18 , Q1 = 8

IQ R = Q3 - Q1 = 18-8 = 10

So, median and IQ R will not need to be changed

4 0

3 years ago

<img src="https://tex.z-dn.net/?f=%20%5Cpi%20%20x%5E%7B2%7D%20" id="TexFormula1" title=" \pi x^{2} " alt=" \pi x^{2} " align="

ANEK [815]

The formule is about area of a circle.

$\pi x^{2} = a$

We need to solve the equation for x. Where x is the radius of the circle.

In order to solve it for x, we need isolate it for x on left side.

So, first we need to get rid pi from left side.

On dividing both sides by pi, we get

$\frac{\pi x^2}\pi}=\frac{a}{\pi}$

$x^2=\frac{a}{\pi}$

Taking square root on both sides, we get

$\sqrt{x^2}=\sqrt{\frac{a}{\pi}}$

$x=\sqrt{\frac{a}{\pi}}$

4 0

3 years ago

The city of Anvil is currently home to 21000 people, and the population has been growing at a continuous rate of 4% per year. Th

Yanka [14]

Answer:

They'll reach the same population in approximately 113.24 years.

Step-by-step explanation:

Since both population grows at an exponential rate, then their population over the years can be found as:

$\text{population}(t) = \text{population}(0)*(1 + \frac{r}{100})^t$

For the city of Anvil:

$\text{population anvil}(t) = 21000*(1.04)^t$

For the city of Brinker:

$\text{population brinker}(t) = 7000*(1.05)^t$

We need to find the value of "t" that satisfies:

$\text{population brinker}(t) = \text{population anvil}(t)\\21000*(1.04)^t = 7000*(1.05)^t\\ln[21000*(1.04)^t] = ln[7000*(1.05)^t]\\ln(21000) + t*ln(1.04) = ln(7000) + t*ln(1.05)\\9.952 + t*0.039 = 8.8536 + t*0.0487\\t*0.0487 - t*0.039 = 9.952 - 8.8536\\t*0.0097 = 1.0984\\t = \frac{1.0984}{0.0097}\\t = 113.24$

They'll reach the same population in approximately 113.24 years.

4 0

3 years ago

Shishir bought 4000 orange at 70 paisa each. But 400 of them were rotten. He sold 2000 oranges at 90 paisa each.If he plans to m

Ipatiy [6.2K]

Answer:

He needs to sell the rest of the oranges at 75 paisa each.

Step-by-step explanation:

Consider the given information that, Shishir bought 4000 orange at 70 paisa each.

Note: 1 rupees = 100 paisa

Thus, 70 paisa = 70/100 rupees = 0.70 rupees

Therefore, the cost price of 4000 oranges is:

4000×0.70 rupees = 2800 rupees

The selling price of 2000 oranges is:

2000×0.90 rupees = 1800 rupees

The number of oranges now Shishir have:

4000 - 2000 - 400 = 1600

He wants to make a profit of RS 200. Thus the selling price of 4000 oranges should be:

2800 rupees + 200 rupees = 3000 rupees

He earned 1800 rupees by selling 2000 oranges at 90 paisa. So, the remaining amount that he needs to make with 1600 oranges is:

3000 rupees - 1800 rupees = 1200 rupees

Therefore, the cost of one orange is:

1600 oranges = 1200 rupees

1 orange = 1200/1600 rupees

1 orange = 0.75 rupees

Hence, he needs to sell the rest of the oranges at 75 paisa each.

4 0

3 years ago