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DerKrebs [107]
3 years ago
13

Can u help me plz now

Mathematics
1 answer:
Elenna [48]3 years ago
3 0

Answer:

1 1/5 miles per hour

Step-by-step explanation:

3/5 = 0.60

1/2= 0.5 hr

0.60 * 2 = 1.2

1.2 mph = 1 1/5

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Evaulate the function at the indicated values h(t)=t+(3/t)i find h(x-1)
alex41 [277]

<span>h<span>(t)</span>=<span>t<span>34</span></span>−3<span>t<span>14</span></span></span>

Note that the domain of h is <span>[0,∞]</span>.

By differentiating,

<span>h'<span>(t)</span>=<span>34</span><span>t<span>−<span>14</span></span></span>−<span>34</span><span>t<span>−<span>34</span></span></span></span>

by factoring out <span>34</span>,

<span>=<span>34</span><span>(<span>1<span>t<span>14</span></span></span>−<span>1<span>t<span>34</span></span></span>)</span></span>

by finding the common denominator,

<span>=<span>34</span><span><span><span>t<span>12</span></span>−1</span><span>t<span>34</span></span></span>=0</span>

<span>⇒<span>t<span>12</span></span>=1⇒t=1</span>

Since <span>h'<span>(0)</span></span> is undefined, <span>t=0</span> is also a critical number.

Hence, the critical numbers are <span>t=0,1</span>.

I hope that this was helpful.

6 0
3 years ago
The amount of time that a customer spends waiting at an airport check-in counter is a random variable with mean 8.0 minutes and
strojnjashka [21]

Answer:

P ( 5 < X < 10 ) = 1

Step-by-step explanation:

Given:-

- Sample size n = 49

- The sample mean u = 8.0 mins

- The sample standard deviation s = 1.3 mins

Find:-

Find the probability that the average time waiting in line for these customers is between 5 and 10 minutes.

Solution:-

- We will assume that the random variable follows a normal distribution with, then its given that the sample also exhibits normality. The population distribution can be expressed as:

                                   X ~ N ( u , s /√n )

Where

                            s /√n = 1.3 / √49 = 0.2143

- The required probability is P ( 5 < X < 10 ) minutes. The standardized values are:

                        P ( 5 < X < 10 ) = P (    (5 - 8) / 0.2143 <  Z  <  (10-8) / 0.2143   )

                                                 = P ( -14.93 < Z < 8.4 )

- Using standard Z-table we have:

                        P ( 5 < X < 10 ) = P ( -14.93 < Z < 8.4 ) = 1        

7 0
3 years ago
What is the solution for the equation: p - 10 = 5
Nookie1986 [14]
The solution is 15.
3 0
3 years ago
Read 2 more answers
The 10th term of an a.p is -27 and the 5th term is -12 .find the first term?​
horsena [70]

Step-by-step explanation:

a_{10} = - 27... (Given) \\\\\therefore a + 9d = - 27...... (1)\\\\a_{5} = - 12... (Given) \\\\\therefore a + 4d = - 12.......(2)\\\\

Subtracting equation (2) from equation (1)

a + 9d-( a + 4d )= - 27-(-12)\\\\\therefore a +9d - a-4d = - 27+12\\\\\therefore 5d = - 15\\\\\therefore d = \frac {- 15}{5}\\\\\huge \orange {\boxed {\therefore d = -3}} \\\\

Substituting d = - 3 in equation (1), we find:

a + 9\times (-3)= - 27\\\\\therefore a - 27 = - 27\\\\\therefore a  = 27- 27\\\\\huge \red {\boxed {\therefore a  = 0}}\\\\

Hence, first term is zero.

3 0
3 years ago
Given the midpoint of (-10,7) and the endpoint of (5,-3), what is the other endpoint?
xxMikexx [17]

(-25,17)


Hope this helps.

4 0
3 years ago
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