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FrozenT [24]
3 years ago
10

The National Weather Service keeps track of the temperature for a given day in a given city. It also keeps track of this informa

tion at certain times during a given day. The following temperatures were recorded on a January day in New York City.Time of Day10 AM12 Noon2 PM4 PM6 PM8 PM10 PMTemperature (F)30353636343028Over which period(s) of time is the average rate of change zero? What, if anything, can you conclude about the actual temperature fluctuation within this period?a.The average rate of change of temperature is 0 from 2 PM to 4 PM; Nothing can be concluded about the actual temperature fluctuation.b.The average rate of change of temperature is 0 from 2 PM to 4 PM and also from 10 AM to 8 PM; Nothing can be concluded about the actual temperature fluctuation.c.The average rate of change of temperature is 0 from 10 AM to 8 PM; Nothing can be concluded about the actual temperature fluctuation.d.The average rate of change of temperature is 0 from 2 PM to 4 PM and also from 10 AM to 8 PM; The conclusion that can be drawn is that the temperature will not change for that time period.
Mathematics
2 answers:
vlada-n [284]3 years ago
6 0

Answer: The average rate of change of temperature is 0 from 2 PM to 4 PM; Nothing can be concluded about the actual temperature fluctuation.

Step-by-step explanation: because from 10 to 8 the change of temperature has change within the hours

8090 [49]3 years ago
5 0
The correct answer for this question is: The <span>period(s) of time where the average rate of changes to zero is </span>"b.The average rate of change of temperature is 0 from 2 PM to 4 PM and also from 10 AM to 8 PM; Nothing can be concluded about the actual temperature fluctuation."
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Which statement is correct?
Tju [1.3M]
What statement? Im missing some information.
5 0
3 years ago
a bag contains eleven counters. five are white. a counter is taken out the bag and is not replaced. a second counter is taken ou
Fed [463]

Answer:

\displaystyle P(A)=\frac{6}{11}

Step-by-step explanation:

<u>Probabilities</u>

The question describes an event where two counters are taken out of a bag that originally contains 11 counters, 5 of which are white.

Let's call W to the event of picking a white counter in any of the two extractions, and N when the counter is not white. The sample space of the random experience is

\Omega=\{WW,WN,NW,NN\}

We are required to compute the probability that only one of the counters is white. It means that the favorable options are

A=\{WN,NW\}

Let's calculate both probabilities separately. At first, there are 11 counters, and 5 of them are white. Thus the probability of picking a white counter is

\displaystyle \frac{5}{11}

Once a white counter is out, there are only 4 of them and 10 counters in total. The probability to pick a non-white counter is now

\displaystyle \frac{6}{10}

Thus the option WN has the probability

\displaystyle P(WN)=\frac{5}{11}\cdot \frac{6}{10}=\frac{30}{110}=\frac{3}{11}

Now for the second option NW. The initial probability to pick a non-white counter is

\displaystyle \frac{6}{11}

The probability to pick a white counter is

\displaystyle \frac{5}{10}

Thus the option NW has the probability

\displaystyle P(NW)=\frac{6}{11}\cdot \frac{5}{10}=\frac{30}{110}=\frac{3}{11}

The total probability of event A is the sum of both

\displaystyle P(A)=\frac{3}{11}+\frac{3}{11}=\frac{6}{11}

\boxed{\displaystyle P(A)=\frac{6}{11}}

7 0
3 years ago
Let X denote the time from the present until the stock market hits 25,000. If X ≤ 1, you win $10, 000X. If X &gt; 1, you win not
const2013 [10]

Answer: $6321

Step-by-step explanation:

Expected value is given as exactly what you might think it means intuitively, the return you can expect for some type of action.

The basic expected value formula is the probability of an event multiplied by the amount of times the event happens:(P(x) * n).

he formula for the Expected Value for a binomial random variable is:

P(x) * X.

X is the number of trials and P(x) is the probability of success.

The formula for calculating the Expected Value where there are multiple probabilities is:

E(X) = ∑X * P(X)

The equation is basically the same, but here you are adding the sum of all the gains multiplied by their individual probabilities instead of just one probability.

Calculating the expected value of winning as follows:

Given,

Fx= e^-x, x is greater or equal to 0

Now, P less than or equal to zero =

[-e^-x] ^1

e^-0 - e^-1

= 1 - e^-1

= 0.6321

and P(x>1) = 1-0.6321= 0.3679

Therefore,Let y= amount of winnings.

Based on the above given information,

y= $10000 0

P(y)= 0.6321 0.3679

The expected value of winning:

E(y) = ΣyP(y)

E(y) = 10000 × 0.6321 + 0 × 0.3679

E(y) = $6321

Therefore, the expected value of winning is $6321

5 0
3 years ago
43. Which function has an x-intercept of 4?
Ulleksa [173]

Answer:

D

Step-by-step explanation:

When f(x) is equal to 0, x=4.

4 0
2 years ago
Decrease 248 by 30%
Snowcat [4.5K]
Okay this equation really says is what is 30% of 248.

So, lets convert 30% to a fraction, 3/10 which is easier to work with.

All you have to do now is get out a calculator and do 248 *3/10 (or .3) and get 74.4

So subtract 74.4 and get

173.6
8 0
3 years ago
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