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Pepsi [2]
2 years ago
7

Solve 2Q + 5s=r for s and show ALL your steps.

Mathematics
2 answers:
kobusy [5.1K]2 years ago
6 0

Answer:

Step-by-step explanation:

Subtract 2Q from both sides

2Q + 5s − 2Q = r − 2Q

Simplify

5s = r − 2Q

Divide both sides by 5

\frac{5s}{5} =\frac{r}{5} -\frac{20}{5}

Simplify

s=\frac{r-20}{5}

 

likoan [24]2 years ago
4 0

Answer:

2q-r= 5s

divide both sides by 5

2q-r/s =5s/5

therefore S= 2q-r/s

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Hope these helps

Step-by-step explanation:

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Write a system of equations and solve
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c+b=7
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c+b=7 --> c=7-b
5c+45b=195

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3 years ago
Stacy goes to the county fair with her friends. The total cost of ride tickets Is given by the equation c = 3.5t, where c is the
murzikaleks [220]
C = 3.5t....when t = 15
c = 3.5(15)
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3 years ago
Find the possible values for s in the inequality 12s – 20 ≤ 50 – 3s – 25.
Margaret [11]
◆ INEQUALITIES ◆

given \: expression \:  \:  -  \\  \\ 12s - 20 \leqslant 50 - 3s - 25 \\  \\ 12s - 20 \leqslant 25 - 3s \\  \\ adding \: 20 \: both \: sides \:  \: , \:  \\ we \: get \:  -  \\  \\ 12 s  \leqslant 45 - 3s \\  \\ 12s + 3s \leqslant 45 \\  \\ 15s \leqslant 45 \\ s \leqslant  \frac{45}{15}  \\  \\ \\    s \leqslant 3 \:  \:  \:  \:  \:  \: ans.
8 0
3 years ago
The proportion of brown M&amp;M's in a milk chocolate packet is approximately 14%. Suppose a package of M&amp;M's typically cont
marshall27 [118]

Answer:

Kindly check explanation

Step-by-step explanation:

Given the scenario above :

A) State the random variable.

The random variable is the p opoertion of brown M&M's in a milk chocolate packet.

B.) Argue that this is a binomial experiment.

Each trial is independent for a total number of 52 trials with a set probability of success at 0.14

C) probability that 6 M&M's are brown:.

P(x) = nCx * p^x * (1-p)^(n-x)

p = 0.14 ; (1 - p) = 0.86 ; n = 52 ; x = 6

P(x = 6) = 52C6 × 0.14^6 × 0.86^46

= 20358520 × 0.00000752954 × 0.00097035078

= 0.1487

D) P(x =25)

P(x = 25) = 52C25 × 0.14^25 × 0.86^27

= 477551179875952 × 449.987958058*10^(-24) × 0.01703955245

= 0.00000000366

E) P(x = 52)

P(x = 52) = 52C52 × 0.14^52 × 0.86^0

= 1 × 3968.78758299*10^(-48) × 1

= 3968.78758299*10^(-48)

F) yes it would be unusual, because such probability is extremely low. However, if a huge or substantial number of trials such may occur

3 0
3 years ago
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