Find the sum. 2y y2 + 2y + 1 +4/ y2 + 3y + 2

A chemist is conducting an experiment which involves a certain chemical reaction. He begins the experiment with 666 milliliters

Ket [755]

Answer:

8 Minutes

Step-by-step explanation:

since we are trying to find how many minutes have passed we first need to find how much substance was used during the time. we can do that by subtracting the leftover substance from the amount we started with.

666-562 = 104

Now that we know how much substance was used we can then use our conversion rate (13 per minute) and find how many minutes have passed by dividing.

104 / 13 = 8

4 0

3 years ago

Write an equation for a function y=square root of x then shifted up 8 units

Brilliant_brown [7]

Answer:

y= x^2 + 8

Step-by-step explanation:

x square rooted is x^2 and shifted up 8 unit probably means that it wants the y-intercept to be 8.

7 0

3 years ago

I need help with this story problem.. 8th Grade Chapter 9.3 Linear Functions:

Paladinen [302]

(1/5)x + 8 This will find 20% of the wholesale cost and add the $8 markup, where 'x' is the wholesale cost

8 0

3 years ago

2. To estimate the mean age for a population of 4000 employees, a simple random sample of 40 employees is selected. a. Would you

ANEK [815]

Answer:

Step-by-step explanation:

a.

Size of the population, N = 4000

Size of the sample, n = 40

n/N = 40/4000 = 0.01

0.01 is less than 0.05 and hence we would not us the finite population correction factor in calculating the standard error of the mean.

b.

Population standard deviation is σ = 8.2

So, the standard error of x’ using the finite population correction factor is give by

σ(x’) = √[(N - n)/(N - 1)] x (sigma/√n)

σ(x’) = √[(4000 - 40)/(4000 - 1)] x (8.2/√40)

σ(x’) = 1.29

Standard error of x’ without using the finite population correction factor is

σ(x’) = σ/√n

σ(x’) = 8.2/√40 = 1.2965

There is little difference between the two values of the standard error. So we can ignore the population correction factor.

c.

Let the population mean be μ

Probability that the sample mean will be within =-2 of the population mean is

P(μ– 2 < x’ < μ + 2)

At x’ = μ – 2 , we have

z = (μ – 2 – μ)/1.2965

z = -1.54

at x’ = μ + 2, we have

z = (μ + 2 – μ)/1.2965

z = 1.54

So the required probability is

P(μ – 2 < x’ < μ + 2) = p(-1.54< z < 1.54)

P(μ – 2 < x’ < μ + 2) = p(z < 1.54) – p(z < -1.54)

P(μ – 2 < x’ < μ + 2) = 0.9382 – 0.0618

P(μ – 2 < x’ < μ + 2) = 0.8764

7 0

3 years ago

One canned orange juice is 25% orange juice another is 5% orange juice. How many liters of each should be mixed together in orde

Cerrena [4.2K]

Answer:

$X = \frac{1.6 - 0.05Y}{0.25}= 6.4 -0.2 Y$ (3)

Replcaing equation (3) into equation (2) we got:

$0.75(6.4 -0.2 Y) +0.95 Y = 18.4$

And solving for Y we got:

$4.8 -0.15 Y +0.95 Y = 18.4$

$0.8 Y = 13.6$

$Y = 17$

And solving for X from equation (3) we got:

$X= 6.4 -0.2*17 = 3$

So we need 3L of orange juice with 25% of concentration and 17 L of orange juice with 5% of concentration

Step-by-step explanation:

For this problem we can work with the concentration of water and orange juice.

Let X the amount for the orange juice with 25% content and Y the amount for the orange juice with 5% of content

Using the concentration of orange juice we have:

$0.25 X + 0.05 Y = 20*0.08$ (1)

And for the water we have:

$0.75 X + 0.95Y = 20*0.92$ (2)

If we solve for X from equation (1) we got:

$X = \frac{1.6 - 0.05Y}{0.25}= 6.4 -0.2 Y$ (3)

Replcaing equation (3) into equation (2) we got:

$0.75(6.4 -0.2 Y) +0.95 Y = 18.4$

And solving for Y we got:

$4.8 -0.15 Y +0.95 Y = 18.4$

$0.8 Y = 13.6$

$Y = 17$

And solving for X from equation (3) we got:

$X= 6.4 -0.2*17 = 3$

So we need 3L of orange juice with 25% of concentration and 17 L of orange juice with 5% of concentration

7 0

3 years ago