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kolezko [41]
3 years ago
12

6.2 x 4.3 + 7.6 simplify

Mathematics
1 answer:
____ [38]3 years ago
4 0
6.2x4.3= 26.66; Add that to 7.6= 34.26; then simplify
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How much money is the sales tax ? I need the answer please
Temka [501]

Answer:

The 3rd one that's what my sis said

5 0
3 years ago
a wall is 8 feet tall and 36 feet wide. tom wants to cover it with white paint that costs 2 dollars per ounce. if each ounce of
arlik [135]

Answer:

1152

Step-by-step explanation:

First we need to find the area of the wall in square inches

8 ft = 8*12 = 96 inch tall

36 ft = 36 * 12 = 432 inches wide

The area = 96 * 432 =41472 in^2

Each ounce of paint cover 72 in^2 so divide by 72

41472/72 =576

We need 576 ounces to cover the wall

At 2 dollars per ounces

576*2 =1152 dollars to paint the wall

8 0
2 years ago
Shannon uses 30 inches of ribbon to make one bow how many feet of ribbon are needed to make 10 bows
coldgirl [10]
First, we need to calculate the amount of ribbon in inches, and then convert it into feet. 

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Hope this helped!
6 0
3 years ago
Read 2 more answers
My supersonic jet burns 4 gallons of jet fuel in 3 seconds. How many gallons of jet fuel do I need to fly for an hour?
Dahasolnce [82]
60 seconds in a minute, but if you divide 60/3, you’ll get 20. So 20*4=80 gallons/min. If you multiply 80*60=4,800 gallons/hr
7 0
2 years ago
The compressive strength of concrete is normally distributed with mu = 2500 psi and sigma = 50 psi. A random sample of n = 8 spe
Nata [24]

Answer:

X \sim N(2500,50)  

Where \mu=2500 and \sigma=50

We select a sample size of n =8. Since the distribution for X is normal then we know that the distribution for the sample mean \bar X is given by:

\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})

And the standard error would be:

SE = \frac{50}{\sqrt{8}}= 17.678 psi

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

Solution to the problem

Let X the random variable that represent the compressive strength of concrete of a population, and for this case we know the distribution for X is given by:

X \sim N(2500,50)  

Where \mu=2500 and \sigma=50

We select a sample size of n =8. Since the distribution for X is normal then we know that the distribution for the sample mean \bar X is given by:

\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})

And the standard error would be:

SE = \frac{50}{\sqrt{8}}= 17.678 psi

7 0
3 years ago
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