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Korvikt [17]
3 years ago
14

At 10:00 a.m. the temperature is –3.4°F. The temperature increases 0.5°F each hour for the next 6 hours. If the temperature drop

s a total of 5.3°F over the following 3 hours, what is the temperature at 7:00 p.m.?
please give an explanation too!
Mathematics
1 answer:
antiseptic1488 [7]3 years ago
6 0

Answer:

-16.3

Step-by-step explanation:

First multiply 0.5 by 6 since the six hours and every hour is 0.5 degrees less or you can add 0.5 six times. You then take the answer which would be 3 and add 3 to -3.4 resulting in -0.4. Afterwards you would take -0.4 and subtract 5.3 degrees three times. This gives you -16.3 degrees. You would be at 7:00 PM because 10:00 a.m. plus six hours would be 4 o'clock and the next 3 hours would add on to 7.

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von has a rectangular workroom with a perimeter of 26 feet. the length of the workroom is 6 feet. what is the width of vons work
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Evaluate the function given the value<br> f(x) = 1 - 2x; f(-12)<br> r
vesna_86 [32]

Answer:

The answer is 25

Step-by-step explanation:

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3 0
3 years ago
Mr. Mole left his burrow and started digging his way down at a constant rate.
Art [367]

Mr. Mole's burrow was at an altitude of 6 meters below the ground.

Step-by-step explanation:

Step 1:

We need to determine the distance that Mr. Mole covers in a single minute.

To do that we divide the difference in values of altitude by the difference in the time periods.

For the first case, Mr. Mole had traveled -18 meters in 5 minutes.

We also have, he traveled -25.2 meters in 8 minutes.

Step 2:

The distance he covered in 1 minute = \frac{-25.2-(-18)}{8-5} = \frac{-25.2+18}{3},

\frac{-25.2+18}{3} = \frac{-7.2}{3} = -2.4.

So with every minute, Mr. Mole digs down an additional 2.4 meters below the surface.

To determine where Mr. Mole's burrow is we subtract the distance traveled in 5 minutes from -18.

The altitude of Mr. Mole's burrow = -18 - 5(-2.4) = -18+12 = -6.

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5 0
3 years ago
Read 2 more answers
A company surveyed 2400 men where 1248 of the men identified themselves as the primary grocery shopper in their household. ​a) E
polet [3.4K]

Answer:

a) With a confidence level of 98%, the percentage of all males who identify themselves as the primary grocery shopper are between 0.4962 and 0.5438.

b) The lower limit of the confidence interval is higher that 0.43, so if he conduct a hypothesis test, he will find that the data shows evidence to said that the fraction is higher than 43%.

c) \alpha =1-0.98=0.02

Step-by-step explanation:

If np' and n(1-p') are higher than 5, a confidence interval for the proportion is calculated as:

p'-z_{\alpha/2}\sqrt{\frac{p'(1-p')}{n} }\leq  p\leq p'+z_{\alpha/2}\sqrt{\frac{p'(1-p')}{n} }

Where p' is the proportion of the sample, n is the size of the sample, p is the proportion of the population and z_{\alpha/2} is the z-value that let a probability of \alpha/2 on the right tail.

Then, a 98% confidence interval for the percentage of all males who identify themselves as the primary grocery shopper can be calculated replacing p' by 0.52, n by 2400, \alpha by 0.02 and z_{\alpha/2} by 2.33

Where p' and \alpha are calculated as:

p' = \frac{1248}{2400}=0.52\\\alpha =1-0.98=0.02

So, replacing the values we get:

0.52-2.33\sqrt{\frac{0.52(1-0.52)}{2400} }\leq  p\leq 0.52+2.33\sqrt{\frac{0.52(1-0.52)}{2400} }\\0.52-0.0238\leq p\leq 0.52+0.0238\\0.4962\leq p\leq 0.5438

With a confidence level of 98%, the percentage of all males who identify themselves as the primary grocery shopper are between 0.4962 and 0.5438.

The lower limit of the confidence interval is higher that 0.43, so if he conduct a hypothesis test, he will find that the data shows evidence to said that the fraction is higher than 43%.

Finally, the level of significance is the probability to reject the null hypothesis given that the null hypothesis is true. It is also the complement of the level of confidence. So, if we create a 98% confidence interval, the level of confidence 1-\alpha is equal to 98%

It means the the level of significance \alpha is:

\alpha =1-0.98=0.02

4 0
3 years ago
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