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3 years ago

A car drove for 50 seconds in that time it traveled 100 meters. What’s the speed of the car?

Mathematics

2 answers:

Marrrta [24]3 years ago

8 0

Answer:

2m/s

2 meters per second

Step-by-step explanation:

Formula for speed is distance (meters) over time (seconds) after you divide you get 2. Please give me the brainliest answer?

:) Hoped this helped!!! Have a good day!!! <3

vesna_86 [32]3 years ago

6 0

Answer:

2 Meters per second

Step-by-step explanation:

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If a cup of coffee has temperature 97Â°C in a room where the ambient air temperature is 25Â°C, then, according to Newton's Law o

FinnZ [79.3K]

Answer:

The average temperature is $T_{a} = 81.95^oC$

Step-by-step explanation:

From the question we are told that

The temperature of the coffee after time t is $T(t) = 25 + 72 e^{[-\frac{t}{45} ]}$

Now the average temperature during the first 22 minutes i.e fro $0 \to 22$ minutes is mathematically evaluated as

$T_{a} = \frac{1}{22-0} \int\limits^{22}_{0} {25 +72 e^{[-\frac{t}{45} ]}} \, dx$

$T_{a} = \frac{1}{22} [25 t + 72 [\frac{e^{[-\frac{t}{45} ]}}{-\frac{1}{45} } ] ] \left| 22} \atop {0}} \right.$

$T_{a} = \frac{1}{22} [25 t - 3240e^{[-\frac{t}{45} ]} ] \left | 45} \atop {{0}} \right.$

$T_{a} = \frac{1}{22} [25 (22) - 3240e^{[-\frac{22}{45} ]} - (- 3240e^{0} )]$

$T_{a} = \frac{1}{22} [550 - 1987.12 + 3240]$

$T_{a} = 81.95^oC$

8 0

2 years ago

A test score of 48.4 on a test having a mean of 66 and a standard deviation of 11. Find the z-score corresponding to the given

Mariulka [41]

Answer:

A. -1.6; not significant

Step-by-step explanation:

The z-score of a data set that is normally distributed with a mean of $\bar x$ and a standard deviation of $\sigma$ , is given by:

$z=\frac{x-\bar x}{\sigma}$ .

From the question, the test score is: $x=48.4$ , the mean is $\bar x=66$ , and the standard deviation is $\sigma =11$ .

We just have to plug these values into the above formula to obtain:

$z=\frac{48.4-66}{11}$ .

This simplifies to: $z=\frac{-17.6}{11}$ .

$z=-1.6$ .

We can see that the z-score falls within two standard deviations of the mean.

Since $-2\le-1.6\le2$ the value is not significant.

The correct answer is A. -1.6; not significant

3 0

3 years ago