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3 years ago

D-13=6 F-26=12 M+18=109 56+k=231 How do you solve these with steps

Mathematics

1 answer:

lapo4ka [179]3 years ago

8 0

Answer:

D=6+13=19

Step-by-step explanation:

F=26+12=38

M=109-18=91

k=231-56=175

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EFGH is a rectangle. EA is 12 in. Find the length of the diagonal EH

Pepsi [2]

Answer:

your answer is 24 in

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3 years ago

Evaluate. 4.3 + 2.75 A. 2.78 B. 3.18 C. 7.05 D. 7.5

miv72 [106K]

1. Set up the long addition.
4.30 + 2.75
2. Calculate 0 + 5, which is 5.
4.30 + 2.75 = 5
3. Calculate 3 + 7, which is 10.
since 10 is two-digit, we carry the first digit 1 to the next column
1
4.30 + 2.75 = 05
1
4.30 + 2.75 = 7.05
5. Therefore, 4.3 + 2.75 = 7.05.
7.05

5 0

3 years ago

Suppose a box of Cracker Jacks contains one of 5 toy prizes: a small rubber ball, a whistle, a Captain America decoder ring, a r

lakkis [162]

Answer:

11.42 boxes

Step-by-step explanation:

For the first box bought, there is a 100% chance of getting a unique toy (since you still don't have any). E₁ = 1.

After that, there is a 4 in 5 chance of getting a unique toy from the next box, the expected number of boxes required is:

$E_2 = (\frac{4}{5})^{-1} = 1.25$

For the next unique toy, there is now a 3 in 5 chance of getting it:

$E_3 = (\frac{3}{5})^{-1} = 1.67$

Following that logic, there is a 2 in 5 chance of getting the 4th unique toy:

$E_4 = (\frac{2}{5})^{-1} = 2.5$

Finally, there is a 1 in 5 chance to get the last unique toy:

$E_5 = (\frac{1}{5})^{-1} = 5$

The expected number of boxes to obtain a full set is:

$E=E_1+E_2+E_3+E_4+E_5\\E=1+1.25+1.67+2.5+5\\E=11.42\ boxes$

5 0

2 years ago

Simplify the first expression 5(m+1)-1=

Olegator [25]

Answer:

-0.8/ -4/5

Step-by-step explanation:

5(m+1) -1 = 0

5m +5 -1 = 0

5m +4 =0

5m= -4

m= -4/5

8 0

3 years ago

Use the Venn diagram to calculate probabilities. Circles A and B overlap. Circle A contains 15, circle B contains 10, and the in

grin007 [14]

Answer:

$(B) P(B)=\dfrac{16}{31}$

Step-by-step explanation:

From the Venn diagram, n(AUB)=31

The given probabilities in the option are calculated below:

$P(A)=\dfrac{21}{31}\\\\P(B)=\dfrac{16}{31}\\\\P(A|B)=\dfrac{P(A\cap B)}{P(B)}= \dfrac{6/31}{16/31} =\dfrac{6}{16}=\dfrac{3}{8} \\\\P(B|A)=\dfrac{P(B\cap A)}{P(A)}= \dfrac{6/31}{21/31} =\dfrac{6}{21}=\dfrac{3}{7}$

The only correct option is the probability of B which is $\frac{16}{31}$

8 0

3 years ago