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3 years ago

Solve for the missing side length Need help please

Mathematics

2 answers:

natita [175]3 years ago

6 0

Answer:

Step-by-step explanation:

Let the missing side be represented as x.

Applying the triangle proportionality theorem, we have:

$\frac{15}{x} = \frac{14 - 4}{4}$

Solve for x

$\frac{15}{x} = \frac{10}{4}$

Cross multiply

$x(10) = 4(15)$

$10x = 60$

Divide both sides by 10

x = 6

babymother [125]3 years ago

3 0

Answer:

Step-by-step explanation:

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A small grocery store had 10 cartons of milk, 2 of which were sour. If you are going to buy the 6th carton of milk sold that day

andre [41]

Answer:

The probability that the sixth customer buys sour milk is $\frac{1}{5}$ .

Step-by-step explanation:

The grocery store has a total of 10 cartons of milk.

The number of cartons of milk that are sour is, 2.

If none of the sour cartons of milk were bought by the first 5 buyers, then the probability of this event is:

P (Both the sour cartons are available to be sold to the sixth customer)

= $P(2\ sour\ cartons)=\frac{2}{5}$

If only one sour carton of milk is sold to the first 5 buyers then the probability is:

P (Only one sour cartons is available to be sold to the sixth customer)

= $P(1\ sour\ cartons)=\frac{1}{5}$

If both the sour carton of milk is sold to the first 5 buyers then the probability is:

P (None of the sour cartons is available to be sold to the sixth customer)

= $P(0\ sour\ cartons)=\frac{0}{5}$

Compute the probability that the sixth customer buys sour milk:

= P (Both sour milk is available for the 6th customer) +

P (Only one sour milk is available for the 6th customer) +

P (None of the sour milk is available for the 6th customer)

$=\frac{{8\choose 5}{2\choose 0}}{{10\choose 5}} \times\frac{2}{5} +\frac{{8\choose 4}{2\choose 1}}{{10\choose 5}} \times\frac{1}{5} +\frac{{8\choose 3}{2\choose 2}}{{10\choose 5}} \times\frac{0}{5} \\=\frac{56\times2}{252\times5} +\frac{140\times1}{252\times5} +0\\=\frac{1}{5}$

Thus, the probability that the sixth customer buys sour milk is $\frac{1}{5}$ .

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