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3 years ago

A dozen eggs cost $1.10 in Dover. In Ensley, the eggs cost 10% more than in Dover. Find the price of a dozen eggs in Ensley.

Mathematics

1 answer:

Julli [10]3 years ago

5 0

1.10 multiplied by .10 = .11
1.10 + .11 = 1.21

In Ensley, the price of a dozen eggs costs $1.21.

Hope this helps.

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There is a 4 digit number that when multiplied by 4 is the reverse.

mote1985 [20]

Answer:

The answer is 2178, if you multiply it by 4, you would get 8717

Step-by-step explanation:

HERE ARE THE STEPS IF YOU NEED THEM!!

1. Lets say the number is written as ABCD (1000a+100b+10c+d).

2.Multiplying it by 4, gives you DCBA (1000d+100c+10b+a)

3. By divisibility of 4, BA is a two-digit number divisible by 4.

4.Now, the largest possible value of ABCD is 2499, because anything more will result in a five digit number when multiplied by 4.

5. Out of the two possibilites 1 and 2 for A, [3] follows that A is even, so 2.

6. From [5], D is either 8 or 9. However, since D times 4 would end in 2, D is 8.

7. Using [2], 4000a+400b+40c+4d = 1000d+100c+10b+a, which would simplify to 2c-13b = 1 after substituting A = 2 and D = 8.

8.The only possible value of (b,c) where they would both be single digit integers is (1,7). Also to note from [3], B is odd. And from [4], B is either 1 or 3.

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3 years ago

Read 2 more answers

Solve for x in the following 4/2.6=5/x

d1i1m1o1n [39]

The value of X is 3.25

Look at the attached picture

Hope it will help you

Good luck on your assignment

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3 years ago

HELP ASAP :

marishachu [46]

4.85×88=$426.80.

1) $426.80 is the answer

7 0

3 years ago

FACTOR THE GCF y(x^2+2)+3(x^2+2)

ioda

$y(x^2+2)+3(x^2+2)=(x^2+2)(y+3)$

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3 years ago

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two telephone calls come into a switchboard at times that are uniformly distributed in a fixed one-hour period. assume that the

AleksandrR [38]

We wil assume a variable x to be the total number of calls received by the switchboard.

The question also says to assume that the calls were made independently.

Given:

Calls are independent.

Calls are uniformly distributed over a 1 hour period.

Ans (a). The calls are distributed uniformly over 1 hour, hence: (0, 1).

So we have,

f1(x1) = 1

f2(x2) = 1

X1 and X2 are considered to be independent of each other. Hence,

f(x1,x2) = f1 (x1) f2 (x2)

f(x1,x2) = 1 (1)

f(x1,x2) = 1

Thus,

P(X1 <= 0.5; X2 <= 0.5) = ∫0.50 ∫0.50 f(x1,x2) dx2 dx1

= ∫^0.5 0 ∫^0.5 0 (1) dx2 dx1

= ∫^0.5 0 (x2^0.5 0) dx1

= ∫^0.5 0 (0.5 - 0) dx1

= 0.5 ∫^0.5 0 dx1

= 0.5 (x1^0.5 0)

= 0.5 (0.5 - 0)

= 0.25

Therefore, the probability that the calls were received within the first 30 minutes or first half hour is 0.25.

Ans (b). Steps 1 and 2 are the same as the above answer.

Probability = [∫^11/12 0 ∫^x1 + 1/12 x1 1dx2 dx1] + [∫^1 1/12 ∫^x1 x1-1/12 1dx2 dx1

= [∫^11/12 0 (x2 ^x1+1/12 x1 dx1] + [∫^1 1/12 (x2 ^x1 x1-1/12 dx1)]

= [∫^11/12 0 (x1 + 1/12 -x1) dx1] + [∫^1 1/12 (x1 - x1 + 1/12) dx1]

= [(1 + x1/12 - 1) ^11/12 0] + [( 1 - 1 + x1/12) ^1 1/12]

= 11/144 + 11/144

= 0.1528

Therefore, the probability that the calls were received within five minutes of each other is 0.15.

#SPJ4

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10 months ago