Find the slope of the line through the given points. (8,0) and (10, 10)

The process that produces sonora bars (a type of candy) is intended to produce bars with a mean weight of 56 gm. the process sta

Nataly [62]

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3 0

3 years ago

A container is filled with 100 grams of bird feed that is 80% seed. Tom and Sally want to mix the 100 grams with bird feed that

marta [7]

Answer:

Initially, we had:

100 grams of a mixture that is 80% seed, (or 0.8 in decimal form)

This means that the quotient between the mass of seeds and the total mass of the mixture is equal to 0.8

Then the mass of seeds, x, can be calculated as:

x/100g = 0.8

x = 0.8*100g

x = 80g

We also have another mixture, that is 5% seed (or 0.05 in decimal form)

This means that in X grams of this mixture, there will be y grams of seeds, such that:

y/X = 0.05

y = 0.05*X grams

Now, we want to create a mixture that is 40% seeds (or 0.4 in decimal form).

Let's do this.

If we add X grams of the 5% mixture, the total mass will be:

M = 100g + X.

The total mass of seeds will be:

m = 80g + X*0.05

and we want that the quotient between the mass of seeds and the total mass to be exactly 0.4

m/M = 0.4

(80g + X*0.05)/(100g + X) = 0.4

Let's solve this for X.

(80g + X*0.05) = 0.4*(100g + X)

80g + X*0.05 = 40g + X*0.4

80g - 40g = X*0.4 - X*0.05

40g = X*0.35

40g/0.35 = X = 114.2g

If we want the mixture to be exactly 40% seeds, we need to add 114.2g of the 5% mixture. (We already can see that Tom is closer to the exact result than Sally)

If we add 114g, like Tom wants, we will have:

M = 100g + 114g = 214g

m = 80g + 0.05*114g = 85.7g

m/M = 85.7g/214g = 0.4005

This is a really good aproximation, if we write this in percentage form this would be:

0.4005*100% = 40.05%

8 0

3 years ago

Which of the following is the same as 2.3 x 10^3? a) .23 b) 230 c) .023 d) 2,300

3241004551 [841]

The answer is D you move the decimal over 1 then add 2 0

7 0

3 years ago

Read 2 more answers

mr.browns salary is 32,000 and imcreases by $300 each year, write a sequence showing the salary for the first five years when wi

chubhunter [2.5K]

Hello!

We have the following data:

a1 (first term or first year salary) = 32000

r (ratio or annual increase) = 300

n (number of terms or each year worked)

We apply the data in the Formula of the General Term of an Arithmetic Progression, to find in sequence the salary increases until it exceeds 34700, let us see:

formula:

$a_n = a_1 + (n-1)*r$

* second year salary

$a_2 = a_1 + (2-1)*300$

$a_2 = 32000 + 1*300$

$a_2 = 32000 + 300$

$\boxed{a_2 = 32300}$

* third year salary

$a_3 = a_1 + (3-1)*300$

$a_3 = 32000 + 2*300$

$a_3 = 32000 + 600$

$\boxed{a_3 = 32600}$

* fourth year salary

$a_4 = a_1 + (4-1)*300$

$a_4 = 32000 + 3*300$

$a_4 = 32000 + 900$

$\boxed{a_4 = 32900}$

* fifth year salary

$a_5 = a_1 + (5-1)*300$

$a_5 = 32000 + 4*300$

$a_5 = 32000 + 1200$

$\boxed{a_5 = 33200}$

We note that after the first five years, Mr. Browns' salary has not yet surpassed 34700, let's see when he will exceed the value:

* sixth year salary

$a_6 = a_1 + (6-1)*300$

$a_6 = 32000 + 5*300$

$a_6 = 32000 + 1500$

$\boxed{a_6 = 33500}$

* seventh year salary

$a_7 = a_1 + (7-1)*300$

$a_7 = 32000 + 6*300$

$a_7 = 32000 + 1800$

$\boxed{a_7 = 33800}$

* eighth year salary

$a_8 = a_1 + (8-1)*300$

$a_8 = 32000 + 7*300$

$a_8 = 32000 + 2100$

$\boxed{a_8 = 34100}$

* ninth year salary

$a_9 = a_1 + (9-1)*300$

$a_9 = 32000 + 8*300$

$a_9 = 32000 + 2400$

$\boxed{a_9 = 34400}$

* tenth year salary

$a_{10} = a_1 + (10-1)*300$

$a_{10} = 32000 + 9*300$

$a_{10} = 32000 + 2700$

$\boxed{a_{10} = 34700}$

we note that in the tenth year of salary the value equals but has not yet exceeded the stipulated value, only in the eleventh year will such value be surpassed, let us see:

* eleventh year salary

$a_{11} = a_1 + (11-1)*300$