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Genrish500 [490]
3 years ago
12

5 cakes cost £4 how much do 7 cakes cost

Mathematics
2 answers:
Shtirlitz [24]3 years ago
7 0

£5.60

divide £4 by 5 to obtain the cost of 1 cake then multiply this value by 7 for the cost of 7 cakes

5 → £4

7 → £4 × \frac{7}{5} = £5.60


Nostrana [21]3 years ago
4 0

You can solve this using proportion.

5 = 4

7 = ?

7(4)/5= 5.6

7 cakes cost £5.6

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The approximate value of log_{8}24 is 1.52832 or 1.53.

What is the logarithmic Expression?

A logarithmic expression is an function that involves the logarithm of an expression containing a variable. To solve exponential equations, first see whether you can write both sides of the equation as powers of the same number.

Here log_{8}24 = log_{2^{3} }24

                   = \frac{1}{3} log_{2}24

                   = \frac{1}{3} log_{2}{8.3}

                   = \frac{1}{3} log_{2}8 + \frac{1}{3} log_{2}3

                   = \frac{1}{3} log_{2}2^3 + \frac{1}{3} log_{2}3

                   = \frac{3}{3} log_{2}2 +\frac{1}{3} log_{2}3

                   =  1 + \frac{1}{3} log_{2}3

                   = 1 + 0.5283

      llog_{8}24  ≈ 1.53

The approximate value of log_{8}24  is 1.52832 or 1.53.

Learn more about Logarithmic expression from :

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4 0
2 years ago
A 12 ft ladder leans against a wall so that the base of the ladder is 5 ft from the wall How high, to the nearest foot, up on th
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Step-by-step explanation:

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3 years ago
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Full Question

Let X be from a geometric distribution with probability of success p.

Given that P ( X > a + b | X > a ) = q ^ { b } = P ( X > b ) for any positive integer x.

Show that for positive integers a and b, P(X > a + b | X > a) = P(X > b).

Answer:

P(X > a + b | X > a) = P(X > b)

Proved --- See Explanation

Step-by-step explanation:

Given

P ( X > a + b | X > a ) = q ^ { b } = P ( X > b )

From the above.

We can derive the following

P(X > a), P(X > b) and P(X > a + b)

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P(X > b) = q^b

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Using the definition of conditional probability

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Bringing both sides together, we're left with

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By substituton

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P(X > a + b | X > a) = q^b

Since P(X > b) = q^b

So,

P(X > a + b | X > a) = P(X > b)

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