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Gnoma [55]
3 years ago
14

E=I(R+r) solve for R

Mathematics
2 answers:
svetlana [45]3 years ago
8 0
Divide I from both sides then subtract r from both sides, too so in the end:

R= (E-r)/I
FinnZ [79.3K]3 years ago
7 0
E÷I =R+r

R=(E÷I) - r
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Quetion to represent each problem the sum of 15 and six times a number tea is 80 One what is the number
andrew-mc [135]

Answer: 11

Step-by-step explanation: The sum of 15 and six times t will be 15 + 6t = 81.

Now, you subtract 15 from both sides to isolate the constants from the variables on the left side and on the other side and you will end up with 6t = 66. Then, you divide 6 from both sides to finally isolate the numbers from the variable and 66 divided by 6 would equal 11


Hope this Helps :)


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4 years ago
Jimmy has a diecast metal car that is a scale model of an actual race car. If the actual length of the car is 10 feet 6 inches a
kirill115 [55]
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8 0
3 years ago
A bag contain 3 black balls and 2 white balls.
Troyanec [42]

Answer:

Step-by-step explanation:

Total number of balls = 3 + 2 = 5

1)

a)

Probability \ of \ taking \ 2 \ black \ ball \ with \ replacement\\\\ = \frac{3C_1}{5C_1} \times \frac{3C_1}{5C_1} =\frac{3}{5} \times \frac{3}{5} = \frac{9}{25}\\\\

b)

Probability \ of \ one \ black \ and \ one\ white \ with \ replacement \\\\= \frac{3C_1}{5C_1} \times \frac{2C_1}{5C_1} = \frac{3}{5} \times \frac{2}{5} = \frac{6}{25}

c)

Probability of at least one black( means BB or BW or WB)

 =\frac{3}{5} \times \frac{3}{5} + \frac{3}{5} \times \frac{2}{5} + \frac{2}{5} \times \frac{3}{5} \\\\= \frac{9}{25} + \frac{6}{25} + \frac{6}{25}\\\\= \frac{21}{25}

d)

Probability of at most one black ( means WW or WB or BW)

=\frac{2}{5} \times \frac{2}{5} + \frac{3}{5} \times \frac{2}{5} \times \frac{2}{5} + \frac{3}{5}\\\\= \frac{4}{25} + \frac{6}{25} + \frac{6}{25}\\\\=\frac{16}{25}

2)

a) Probability both black without replacement

  =\frac{3}{5} \times \frac{2}{4}\\\\=\frac{6}{20}\\\\=\frac{3}{10}

b) Probability  of one black and one white

 =\frac{3}{5} \times \frac{2}{4}\\\\=\frac{6}{20}\\\\=\frac{3}{10}

c) Probability of at least one black ( BB or BW or WB)

 =\frac{3}{5} \times \frac{2}{4} + \frac{3}{5} \times \frac{2}{4} + \frac{2}{5} \times \frac{3}{4}\\\\=\frac{6}{20} + \frac{6}{20} + \frac{6}{20} \\\\=\frac{18}{20} \\\\=\frac{9}{10}

d) Probability of at most one black ( BW or WW or WB)

 =\frac{3}{5} \times \frac{2}{4} + \frac{2}{5} \times \frac{1}{4} + \frac{2}{5} \times \frac{3}{4}\\\\=\frac{6}{20} + \frac{2}{20} + \frac{6}{20} \\\\=\frac{14}{20}\\\\=\frac{7}{10}

6 0
3 years ago
Graph: Y – 3=1/2(x +2)
tatuchka [14]

Answer:

Step-by-step explanation:y–3=½(x+2)

Multiply through by the denominator (2).

2×y–2×3=½(x+2)×2

2y–6=1(x+2) because the denominator (2) will cancel the other 2 multiplying it

2y–6=x+2

2y=x+2+6

2y=x+8

---. -----

2. 2

•°•y=x+8

-----. Or y=x+4

2

5 0
3 years ago
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