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3 years ago

15

I am going to mix a gallon bucket of window cleaner. The instruction direct you mix 1 part cleaner to 3 parts water. How much cl

eaning solution will I need to use

1 answer:

Darya [45]3 years ago

7 0

1 quart bc there will be one quart of the solution to 3 quarts water, equalling a gallon

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Rewrite as a simplefied fraction

9966 [12]

Answer:
79/25

Work:

Firstly, you have to change the decimal portion of 3.16 (0.16) into a fraction, then you simplify it by a common factor, which is 4. You then convert your whole number, which is 3, into an improper fraction by multiplying it by 25 and making that the numerator, 25 being the denominator of the fraction. Then, you add the two fractions and voila! You have a simplified fraction!

0.16 = 16/100

16/100 = 4/25

3 x 25 = 75

75/25 + 4/25 = 79/25

3 0

3 years ago

For the month of November in a certain city, 88% of the days are cloudy. Also in the month of November in the same city, 54%

Gnoma [55]

Answer:

There is a 61.36% probability that a randomly selected day in November will be foggy if it is cloudy.

Step-by-step explanation:

We have these following probabilities:

An 88% probability that the day is cloudy.

An 54% probability that the day is both foggy and cloudy.

What is the probability that a randomly selected day in November will be foggy if it is cloudy?

This is the percentage of days that are cloudy and foggy divided by those that are cloudy. So:

$P = \frac{0.54}{0.88} = 0.6136$

There is a 61.36% probability that a randomly selected day in November will be foggy if it is cloudy.

4 0

3 years ago

The probability that an American CEO can transact business in a foreign language is .20. Twelve American CEOs are chosen at rand

NNADVOKAT [17]

Answer with Step-by-step explanation:

We are given that

The probability that an American CEO can transact business in foreign language=0.20

The probability than an American CEO can not transact business in foreign language= $1-0.20=0.80$

Total number of American CEOs chosen=12

a. The probability that none can transact business in a foreign language= $12C_0(0.20)^0(0.80)^{12}$

Using binomial theorem $nC_r(1-p)^{n-r}p^r$

The probability that none can transact business in a foreign language= $\frac{12!}{0!(12-0)!}(0.8)^{12}=(0.8)^{12}$

b.The probability that at least two can transact business in a foreign language= $1-P(x=0)-p(x=1)=1-((0.8)^{12}+12C_1(0.8)^{11}(0.2))=1-((0.8)^{12}+12(0.8)^{11}}(0.2))$

c.The probability that all 12 can transact business in a foreign language= $12C_{12}(0.8)^0(0.2)^{12}$

The probability that all 12 can transact business in a foreign language= $\frac{12!}{12!}(0.2)^{12}=(0.2)^{12}$

5 0

2 years ago

I forgot how to do this, can someone help me-

Artyom0805 [142]

Step-by-step explanation:

1) x/20 = -9

By cross multiplication,

x = -180

2) -168 = -14k

k = -168/-14

k = 12

3) -5m = -50

m = 10

4) -320 = -20n

n = 16

5) -4 = x - 5

x = 1

6) -6(-8 + n) = -54

42 - 6n = -54

-6n = -54 -52

6n = 106

n = 17.6

7) 3(x - 9) = -3

3x - 27 = -3

3x = 24

x = 8

8) 9k + 3 = -78

9k = -81

k = -9

9)4 + (n/4) = 8

By taking LCM,

16 + n/4 = 8

16 + n = 32

n = 16

10) -5r + 7 = -73

-5r = -80

r = 16

<em>Hope</em><em> </em><em>it</em><em> </em><em>helps</em><em>.</em>

6 0

1 year ago

2+2 all wrong answers only

cupoosta [38]

Answer: the answer is 1000years of death!!!

Step-by-step explanation:

7 0

2 years ago

Read 2 more answers