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3 years ago

Hi could someone just help me with this I'm not sure if i'm doing it correctly.

1 answer:

4 0

Two angles are supplementary when they add up to 180 degrees ⇒
m∠1 + m∠2 = 180°
4y + 7 + 9y + 4 = 180
13y + 11 = 180
13y = 180 - 11
13y = 169
y = 169/13
y = 13

m∠2 = 9y + 4 = 9 * 13 + 4 = 121°

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The movement of the progress bar may be uneven because questions can be worth more or less (including zero) depending on your an

QveST [7]

Answer:

Three and two I believe

Step-by-step explanation:

6 0

2 years ago

If an 8 hour work day includes a total break time of 22 1/4%, how many hours are used for breaks?

torisob [31]

Answer:

1 39/50 hours. This is equivalent to 1.78 hours.

Step-by-step explanation:

The sum of the hours of work and the hours for break gives the work day hour which has been given as 8 hours.

Furthermore, given that the break time is 22 1/4%, the number of hours for break

= 22 1/4% × 8 hours

= 89/400 × 8

= 1 39/50 hours

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

Since there are no replacement, the probability of taking a second non 75-W bulb is now $\frac{8}{14}=0.5714$

Following this procedure 5 times, we find the probabilities

$\frac{9}{15},\frac{8}{14},\frac{7}{13},\frac{6}{12},\frac{5}{11}$

which are

0.6, 0.5714, 0.5384, 0.5, 0.4545

As the events are independent, the probability of choosing 5 non 75-W bulbs is the product

0.6*0.5714*0.5384*0.5*0.4545 = 0.0419 = 4.19%

3 0

3 years ago

Seven times a number is equal to 9 more than 4 times the number. find the number

Yanka [14]

Represent 'a number' by x
7 times x equals 9 more than 4 times x
7 times x=9+4 times x
7x=9+4x
subtract 4x from both sides
3x=9
divide 3
x=3

the number is 3

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3 years ago

John spends 22% of his monthly budget on his car payment, which is $400. Which proportion can be used to calculate John’s total

nikitadnepr [17]

Answer:

...

5 days ago — John spends 22% of his monthly budget on his car payment, which is $400. Which proportion can be used to calculate John's total monthly

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3 years ago