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3 years ago

Hat is the length of AB and AC

Mathematics

1 answer:

lawyer [7]3 years ago

5 0

Answer:

AB = 17, AC = $\sqrt{433}$

Step-by-step explanation:

Pythagorean's Theorem; A^2 + B^2 = C^2. C is the longest side.

26 - 11 = 15

15^2 + 8^2 = 289

$\sqrt{289} = 17$

$15^2 + 8^2 = 17^2\\$

17^2 + 12^2 = 433

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What is the subset of 2/3??

sineoko [7]

It belongs to the rational numbers.

It is the ratio of two integers, 2 and 3.

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3 years ago

Jean needs to rent a car for an upcoming buiness trip. The cost per day for the car is $30, and the one-time-refueling cost upon

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2 years ago

PLEASE HELP!!!!

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$\displaystyle\\
\bf\\
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3 years ago

Read 2 more answers

Jonathan took 33 pictures on the class trip and 67

Semenov [28]

Answer:

<em>JONATHAN WILL NEED 14 PAGES.</em>

Explanation:

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3 0

3 years ago

Suppose you just received a shipment of nine televisions. Three of the televisions are defective. If two televisions are randoml

Temka [501]

<h2>Answer:</h2>

The probability that both televisions work is: 0.42

The probability at least one of the two televisions does not work is:

0.5833

<h2>Step-by-step explanation:</h2>

There are a total of 9 televisions.

It is given that:

Three of the televisions are defective.

This means that the number of televisions which are non-defective are:

9-3=6

The probability that both televisions work is calculated by:

$=\dfrac{6_C_2}{9_C_2}$

( Since 6 televisions are in working conditions and out of these 6 2 are to be selected.

and the total outcome is the selection of 2 televisions from a total of 9 televisions)

Hence, we get:

$=\dfrac{\dfrac{6!}{2!\times (6-2)!}}{\dfrac{9!}{2!\times (9-2)!}}\\\\\\=\dfrac{\dfrac{6!}{2!\times 4!}}{\dfrac{9!}{2!\times 7!}}\\\\\\=\dfrac{5}{12}\\\\\\=0.42$

The probability at least one of the two televisions does not work:

Is equal to the probability that one does not work+probability both do not work.

Probability one does not work is calculated by:

$=\dfrac{3_C_1\times 6_C_1}{9_C_2}\\\\\\=\dfrac{\dfrac{3!}{1!\times (3-1)!}\times \dfrac{6!}{1!\times (6-1)!}}{\dfrac{9!}{2!\times (9-2)!}}\\\\\\=\dfrac{3\times 6}{36}\\\\\\=\dfrac{1}{2}\\\\\\=0.5$

and the probability both do not work is:

$=\dfrac{3_C_2}{9_C_2}\\\\\\=\dfrac{1}{12}\\\\\\=0.0833$

Hence, Probability that atleast does not work is:

0.5+0.0833=0.5833

4 0

3 years ago