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3 years ago

How long does it take to drive 175 miles at a speed of 25 mph

Mathematics

2 answers:

Ivenika [448]3 years ago

7 0

You multiply miles by speed 175 x 25 = 4375
4375/60 =
Approx 73 minutes
73-60= 1 hour and 13 minutes approx

kotykmax [81]3 years ago

6 0

Speed= distance/time

Therefore, time= distance/speed.

Plug in the known values
t= 175/25
t= 7

Final answer: 7 hrs

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Harlene tosses two number cubes. If a sum of 8 or 12 comes up, she gets 9 points. If not, she loses 2 points. What is the expect

Amanda [17]

P(8 or 12) = 1/6
P(not 8 or 12 = 5/6
Let the expected points fro one roll be X.
$E(X)=9\times\frac{1}{6}-2\times\frac{5}{6}=-\frac{1}{6}\ points$
The answer is -(1/6) points.

5 0

3 years ago

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Please answer 3(x-1)=5x+3-2x

Setler [38]

Answer: there are no solutions

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3
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3
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Step 2: Subtract 3x from both sides.
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6 0

3 years ago

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Show your solution follow the steps need an answer, please need an answer, please.

Alex777 [14]

Answer:

after calculating I got this answer.

hope this answer helps you dear...take care

8 0

3 years ago

Find the vectors T, N, and B at the given point. r(t) = < t^2, 2/3t^3, t >, (1, 2/3 ,1)

maxonik [38]

Answer with Step-by-step explanation:

We are given that

$r(t)=< t^2,\frac{2}{3}t^3,t >$

We have to find T,N and B at the given point t > (1, $2/3$ ,1)

$r'(t)=$

$\mid r'(t) \mid=\sqrt{(2t)^2+(2t^2)^2+1}=\sqrt{(2t^2+1)^2}=2t^2+1$

$T(t)=\frac{r'(t)}{\mid r'(t)\mid}=\frac{}{2t^2+1}$

Now, substitute t=1

$T(1)=\frac{}{2+1}=\frac{1}{3}$

$T'(t)=\frac{-4t}{(2t^2+1)^2} +\frac{1}{2t^2+1}$

$T'(1)=-\frac{4}{9}+\frac{1}{3}$

$T'(1)=\frac{1}{9}=$

$\mid T'(1)\mid=\sqrt{(\frac{-2}{9})^2+(\frac{4}{9})^2+(\frac{-4}{9})^2}=\sqrt{\frac{36}{81}}=\frac{2}{3}$

$N(1)=\frac{T'(1)}{\mid T'(1)\mid}$

$N(1)=\frac{}{\frac{2}{3}}=$

$N(1)=$

$B(1)=T(1)\times N(1)$

$B(1)=\begin{vmatrix}i&j&k\\\frac{2}{3}&\frac{2}{3}&\frac{1}{3}\\\frac{-1}{3}&\frac{2}{3}&\frac{-2}{3}\end{vmatrix}$

$B(1)=i(\frac{-4}{9}-\frac{2}{9})-j(\frac{-4}{9}+\frac{1}{3})+k(\frac{4}{9}+\frac{2}{9})$

$B(1)=-\frac{2}{3}i+\frac{1}{3}j+\frac{2}{3}k$

$B(1)=\frac{1}{3}$

5 0

3 years ago

A store selling newspapers orders only n = 4 of a certain newspaper because the manager does not get many calls for that publica

umka2103 [35]

Answer:

a) The expected value is 2.680642

b) The minimun number of newspapers the manager should order is 6.

Step-by-step explanation:

a) Lets call X the demanded amount of newspapers demanded, and Y the amount of newspapers sold. Note that 4 newspapers are sold when at least four newspaper are demanded, but it can be more than that.

X is a random variable of Poisson distribution with mean $\mu = 3$ , and Y is a random variable with range {0, 1, 2, 3, 4}, with the following values

PY(k) = PX(k) = ε^(-3)*(3^k)/k! for k in {0,1,2,3}
PY(4) = 1 -PX(0) - PX(1) - PX(2) - PX (3)

we obtain:

PY(0) = ε^(-3) = 0.04978..

PY(1) = ε^(-3)*3^1/1! = 3*ε^(-3) = 0.14936

PY(2) = ε^(-3)*3^2/2! = 4.5*ε^(-3) = 0.22404

PY(3) = ε^(-3)*3^3/3! = 4.5*ε^(-3) = 0.22404

PY(4) = 1- (ε^(-3)*(1+3+4.5+4.5)) = 0.352768

E(Y) = 0*PY(0)+1*PY(1)+2*PY(2)+3*PY(3)+4*PY(4) = 0.14936 + 2*0.22404 + 3*0.22404+4*0.352768 = 2.680642

The store is expected to sell 2.680642 newspapers

b) The minimun number can be obtained by applying the cummulative distribution function of X until it reaches a value higher than 0.95. If we order that many newspapers, the probability to have a number of requests not higher than that value is more 0.95, therefore the probability to have more than that amount will be less than 0.05

we know that FX(3) = PX(0)+PX(1)+PX(2)+PX(3) = 0.04978+0.14936+0.22404+0.22404 = 0.647231

FX(4) = FX(3) + PX(4) = 0.647231+ε^(-3)*3^4/4! = 0.815262

FX(5) = 0.815262+ε^(-3)*3^5/5! = 0.91608

FX(6) = 0.91608+ε^(-3)*3^6/6! = 0.966489

So, if we ask for 6 newspapers, the probability of receiving at least 6 calls is 0.966489, and the probability to receive more calls than available newspapers will be less than 0.05.

I hope this helped you!

8 0

3 years ago