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3 years ago

Please hurry

Mathematics

1 answer:

kvv77 [185]3 years ago

5 0

Answer:

if the two angles/two segments are supposed to be set = to each other, then

3y=6y-12

-3y=-12

y=4

7x+5=5x+9

2x=4

x=2

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FAST ASAP!! WILL GIVE BRAINLIST ALOT OF POINTS

Alik [6]

Answer:

= 3 3/4 m^2

or 15/4 m^2

Step-by-step explanation:

The base is the triangle.

A =1/2 bh

A = 1/2 (3 1/3 * 9/4)

Change 3 1/3 to an improper fraction (3*3+1)/3 = 10/3

A = 1/2 * 10/3 *9/4

Re arranging

= 1/2* 9/3 * 10/4

= 1/2 (3) 5/2

=15/4

Now writing as a mixed number

= 3 3/4 m^2

3 0

3 years ago

What is the area of the composite shape

FrozenT [24]

Answer:

Step-by-step explanation:

The shape can be divided into two parts, let use the division in the question

So we have two rectangles

The area of a rectangle is given by l×b.

Let the big rectangle be 1

And the smaller rectangle be 2

Rectangle 1

Length is 16cm, breadth is 6cm

Area=l ×b

Area=16×6

Area=96cm^2

Rectangle two

The length will be 14-6=8cm

And the breadth will be 16-2=14cm

Then,

Area=l ×b

Area=14×8

Area=112cm^2

Total area is

Area1+Area2

112+96

Total area=208cm^2.

5 0

3 years ago

. 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 $\frac{9}{15}=0.6$