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2 years ago

Find the length of AB

Mathematics

1 answer:

Vlad [161]2 years ago

3 0

Answer:

C. 44.98

Step-by-step explanation:

Hi there!

We are given the right triangle ABC, m<B=12°, and CB =44

We want to find the length of AB

We can use trigonometry to do it

Let's find the ratio in reference to angle B, as that angle is given.

In reference to angle B the opposite angle is AC, the adjacent side is CB, and the hypotenuse is AB

Now let's recall the 3 most commonly used functions:

$sine=\frac{opposite}{hypoptenuse}$

$cosine=\frac{adjacent}{hypotenuse}$

$tangent=\frac{opposite}{adjacent}$

Let's find the cosine of angle B, as it uses CB and AB, which are the given side and the side we need to find

In that case,

cos(12)= $\frac{CB}{AB}$

cos(12)= $\frac{44}{AB}$

Multiply both sides by AB

$AB$ *cos(12)=44

Divide both sides by cos(12)

AB= $\frac{44}{cos(12)}$

Now plug $\frac{44}{cos(12)}$ into your calculator. Make sure your calculator is on degree mode

AB≈44.98

So the answer is C

Hope this helps!

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. A box in a certain supply room contains four 40-W lightbulbs, five 60-W bulbs, and six 75-W bulbs. Suppose that three bulbs ar

yaroslaw [1]

Answer:

a) 59.34%

b) 44.82%

c) 26.37%

d) 4.19%

Step-by-step explanation:

(a)

There are in total 4+5+6 = 15 bulbs. If we want to select 3 randomly there are K ways of doing this, where K is the combination of 15 elements taken 3 at a time

$K=\binom{15}{3}=\frac{15!}{3!(15-3)!}=\frac{15!}{3!12!}=\frac{15.14.13}{6}=455$

As there are 9 non 75-W bulbs, by the fundamental rule of counting, there are 6*5*9 = 270 ways of selecting 3 bulbs with exactly two 75-W bulbs.

So, the probability of selecting exactly 2 bulbs of 75 W is

$\frac{270}{455}=0.5934=59.34\%$

(b)

The probability of selecting three 40-W bulbs is

$\frac{4*3*2}{455}=0.0527=5.27\%$

The probability of selecting three 60-W bulbs is

$\frac{5*4*3}{455}=0.1318=13.18\%$

The probability of selecting three 75-W bulbs is

$\frac{6*5*4}{455}=0.2637=26.37\%$

Since the events are disjoint, the probability of taking 3 bulbs of the same kind is the sum 0.0527+0.1318+0.2637 = 0.4482 = 44.82%

(c)

There are 6*5*4 ways of selecting one bulb of each type, so the probability of selecting 3 bulbs of each type is

$\frac{6*5*4}{455}=0.2637=26.37\%$

(d)

The probability that it is necessary to examine at least six bulbs until a 75-W bulb is found, supposing there is no replacement, is the same as the probability of taking 5 bulbs one after another without replacement and none of them is 75-W.

As there are 15 bulbs and 9 of them are not 75-W, the probability a non 75-W bulb is $\frac{9}{15}=0.6$