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Feliz [49]
3 years ago
15

A painter mixes 2 1/2 pints of yellow paint with 4 pints of red paint to make a certain shade of orange paint. How many pints of

yellow paint should be mixed with 10 pints of red paint to make this shade of orange?
Mathematics
2 answers:
9966 [12]3 years ago
7 0

Answer:

6

Step-by-step explanation:

A painter mixes 2 1/2 pints (2.5 pints) of yellow paint with 4 pints of red paint to make a certain shade of orange paint. The ratio of yellow paint to red paint is 2.5:4. In order to find the pints of yellow paint needed to be mixed with 10 pints of red paint, we will use conversion fractions.

10 red pints × (2.5 yellow pints/ 4 red pints) ≈ 6 pints (we round off to 1 significant figure)

Natasha2012 [34]3 years ago
4 0

For every 4 pints of red, there are 2 1/12 pints of yellow. 8 of those red pints would make 5 yellow pints. Adding 2 more pints of red to get to the 10 would mean you have to half the 2 1/2, which would be 1 1/2. So, you would need 6 1/2 pints of yellow mixed with the 10 red pints to make the shade of orange. :)

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For the following telescoping series, find a formula for the nth term of the sequence of partial sums {Sn}. Then evaluate limn→[
Ivenika [448]

Answer:

The following are the solution to the given points:

Step-by-step explanation:

Given value:

1) \sum ^{\infty}_{k = 1} \frac{1}{k+1} - \frac{1}{k+2}\\\\2) \sum ^{\infty}_{k = 1} \frac{1}{(k+6)(k+7)}

Solve point 1 that is \sum ^{\infty}_{k = 1} \frac{1}{k+1} - \frac{1}{k+2}\\\\:

when,

k= 1 \to  s_1 = \frac{1}{1+1} - \frac{1}{1+2}\\\\

                  = \frac{1}{2} - \frac{1}{3}\\\\

k= 2 \to  s_2 = \frac{1}{2+1} - \frac{1}{2+2}\\\\

                  = \frac{1}{3} - \frac{1}{4}\\\\

k= 3 \to  s_3 = \frac{1}{3+1} - \frac{1}{3+2}\\\\

                  = \frac{1}{4} - \frac{1}{5}\\\\

k= n^  \to  s_n = \frac{1}{n+1} - \frac{1}{n+2}\\\\

Calculate the sum (S=s_1+s_2+s_3+......+s_n)

S=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+.....\frac{1}{n+1}-\frac{1}{n+2}\\\\

   =\frac{1}{2}-\frac{1}{5}+\frac{1}{n+1}-\frac{1}{n+2}\\\\

When s_n \ \ dt_{n \to 0}

=\frac{1}{2}-\frac{1}{5}+\frac{1}{0+1}-\frac{1}{0+2}\\\\=\frac{1}{2}-\frac{1}{5}+\frac{1}{1}-\frac{1}{2}\\\\= 1 -\frac{1}{5}\\\\= \frac{5-1}{5}\\\\= \frac{4}{5}\\\\

\boxed{\text{In point 1:} \sum ^{\infty}_{k = 1} \frac{1}{k+1} - \frac{1}{k+2} =\frac{4}{5}}

In point 2: \sum ^{\infty}_{k = 1} \frac{1}{(k+6)(k+7)}

when,

k= 1 \to  s_1 = \frac{1}{(1+6)(1+7)}\\\\

                  = \frac{1}{7 \times 8}\\\\= \frac{1}{56}

k= 2 \to  s_1 = \frac{1}{(2+6)(2+7)}\\\\

                  = \frac{1}{8 \times 9}\\\\= \frac{1}{72}

k= 3 \to  s_1 = \frac{1}{(3+6)(3+7)}\\\\

                  = \frac{1}{9 \times 10} \\\\ = \frac{1}{90}\\\\

k= n^  \to  s_n = \frac{1}{(n+6)(n+7)}\\\\

calculate the sum:S= s_1+s_2+s_3+s_n\\

S= \frac{1}{56}+\frac{1}{72}+\frac{1}{90}....+\frac{1}{(n+6)(n+7)}\\\\

when s_n \ \ dt_{n \to 0}

S= \frac{1}{56}+\frac{1}{72}+\frac{1}{90}....+\frac{1}{(0+6)(0+7)}\\\\= \frac{1}{56}+\frac{1}{72}+\frac{1}{90}....+\frac{1}{6 \times 7}\\\\= \frac{1}{56}+\frac{1}{72}+\frac{1}{90}+\frac{1}{42}\\\\=\frac{45+35+28+60}{2520}\\\\=\frac{168}{2520}\\\\=0.066

\boxed{\text{In point 2:} \sum ^{\infty}_{k = 1} \frac{1}{(n+6)(n+7)} = 0.066}

8 0
3 years ago
When two dice are rolled, what is the probability the two numbers will have a sum of 5?
ira [324]

When two dice are rolled, then possible outcomes are

(1,1) (1,2) (1,3) (1,4) (1,5) (1,6)

(2,1) (2,2) (2,3) (2,4) (2,5) (2,6)

(3,1) (3,2) (3,3) (3,4) (3,5) (3,6)

(4,1) (4,2) (4,3) (4,4) (4,5) (4,6)

(5,1) (5,2) (5,3) (5,4) (5,5) (5,6)

(6,1) (6,2) (6,3) (6,4) (6,5) (6,6)

Total possible outcomes = 36

In that, we count the ordered pairs that give us sum 5

(1,4) (2,3) (3,2) and (4,1) = 4 outcomes

Probability = (number of times the event occurs)/ (total number of outcomes)

P( sum of 5) = \frac{4}{36} = \frac{1}{9}

7 0
3 years ago
a garden table and a bench cost $1,025 combined the garden table cost $75 more than the bench what is the cost of the bench
lukranit [14]
It costs 1,100
you just add the two
7 0
3 years ago
How many numbers lie between the square of 56 and 57
NARA [144]

Answer:

Given, two numbers 56 and 57 we need to find out the numbers lie between the squares of the given numbers.

Now, we have numbers lying between the square of n and (n + 1) is 2n

⇒ Numbers between squares of 56 and (56 + 1) = 2 × 56 = 112

Hence, 112 numbers lies between the square of the given numbers.

7 0
2 years ago
Read 2 more answers
Carbon-14 dating assumes that the carbon dioxide on Earth today has the same radioactive content as it did centuries ago. If thi
defon

7978 years

Step-by-step explanation:

A = A02^(-t/hl)

where hl = half-life.

Dividing both sides by A0 and taking the logarithm, we get

ln(A/A0) = -(t/hl)ln2

or solving for t,

t= -(hl)ln(A/A0)/ln2

note that A/A0 = 0.38

t = -(5715 yrs)[ln(0.38)/ln2]

= 7978 years

7 0
2 years ago
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