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Ksivusya [100]
3 years ago
12

Convert 5.764764764 to a rational expression

Mathematics
1 answer:
NISA [10]3 years ago
8 0

Answer: 5 and 764 over 999

5764 over 9999

5 and 764 over 99

5 and 999 over 764

Step-by-step explanation:

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Write a postive or negative integer that repesents the situation. A football team loses 3 yards
Andreas93 [3]

Answer: -3

Step-by-step explanation:

Loses means negative. It is getting lesser and if you use a number line and start from zero, you move to the left side of it and get -3. I hope that helps.

3 0
3 years ago
DJ has 16 movies, and Three-eighths of them are comedies. How many movies are comedies?
IrinaVladis [17]

Answer:

16 * 3/8 = 6

6 movies are comedies

7 0
3 years ago
Read 2 more answers
How do you solve -1/12 - (-11/12)
valentinak56 [21]

-1/12 - (-11/12) would change into -1/12 + 11/12 because a double negative (bold print) would become a positive. The answer to the new problem would be 10/12 or 5/6 (depending on if it says to simplify).

7 0
3 years ago
A sample of 1500 computer chips revealed that 32% of the chips do not fail in the first 1000 hours of their use. The company's p
Grace [21]

Answer:

Yes, we have sufficient evidence at the 0.02 level to support the company's claim.

Step-by-step explanation:

We are given that a sample of 1500 computer chips revealed that 32% of the chips do not fail in the first 1000 hours of their use. The company's promotional literature claimed that above 29% do not fail in the first 1000 hours of their use.

Let Null Hypothesis, H_0 : p \leq 0.29  {means that less than or equal to 29% do not fail in the first 1000 hours of their use}

Alternate Hypothesis, H_1 : p > 0.29  {means that more than 29% do not fail in the first 1000 hours of their use}

The test statics that will be used here is One-sample proportions test;

          T.S. = \frac{\hat p -p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }  ~ N(0,1)

where, \hat p = proportion of chips that do not fail in the first 1000 hours of their use = 32%

            n = sample of chips = 1500

So, <u>test statistics</u> = \frac{0.32-0.29}{\sqrt{\frac{0.32(1-0.32)}{1500} } }

                              = 2.491

<em>Now, at 0.02 level of significance the z table gives critical value of 2.054. Since our test statistics is more than the critical value of z so we have sufficient evidence to reject null hypothesis as it fall in the rejection region.</em>

Therefore, we conclude that more than 29% do not fail in the first 1000 hours of their use which means we have sufficient evidence at the 0.02 level to support the company's claim.

7 0
3 years ago
ank A contains 80 gallons of water in which 20 pounds of salt has been dissolved. Tank B contains30 gallons of water in which 5
adell [148]

Answer:

\frac{dx}{dt}=2-\frac{3}{40}x+\frac{y}{15}

\frac{dy}{dt}=\frac{3}{40}x-\frac{y}{5}

x(0)=20,y(0)=5

Step-by-step explanation:

We are given that

Tank A contains water=80 gallons

x(0)=20,y(0)=5

Tank B contains water=30 gallons

Rate=4 gallon/min

Concentration of salt  is pumped into tank=0.5 pound /gallon of water

Solution pumped from tank A to tank B at the rate=6 gallons/min

Solution pumped from tank B to tank A at the rate=2gallon/min

Solution from tank B is pumped out of the system at the rate=4 gallon/min

We have to find the DE at time t

For x

Rate in=0.5\times 4+y(t)/30\times 2

Rate in=2+y/15

Rate out=x/80\times 6

Rate out=3x/40[/tex]

\frac{dx}{dt}=Rate in-Rate out

\frac{dx}{dt}=2+y/15-3x/40

\frac{dx}{dt}=2-\frac{3}{40}x+\frac{y}{15}

For y

Rate in=x/80\times 6

Rate in=3/40x

Rate out=y/30\times (4+2)

Rate out=y/5

\frac{dy}{dt}=Rate in-Rate out

\frac{dy}{dt}=\frac{3}{40}x-\frac{y}{5}

8 0
3 years ago
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