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Alex
3 years ago
9

Complete the statement. Round to the nearest hundredth, if necessary.

Mathematics
2 answers:
Scilla [17]3 years ago
8 0

Answer:

0,02\:mi/s ≈ 68\:mph

Step-by-step explanation:

Since there 3600 seconds in an hour, and because you are going from a smaller unit to a bigger unit, you divide 68 by 3600 to get this:

_

0,018

The "8" tells the hundredths digit to round up one, so you approximately get 0,02.

* That bar over the "8" tells you that it repeats.

I am joyous to assist you anytime.

stepladder [879]3 years ago
4 0

Answer:

0.6 mi

Step-by-step explanation:

68     1h           1 min                 0.6mi

__________________  =     ______

1h      60mins      60 sec           1 sec

68

___   = 0.56666666666 rounded 0.6

120

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In 1990, the cost of tuition at a large Midwestern university was $99 per credit hour. In 2003, tuition had risen to $268 per cr
vodka [1.7K]

Answer:

63%

Step-by-step explanation:

<em>From the question, we aim to find the percent increase in the tuition</em>

Given data

initial cost= $99 per credit hour

Final cost=  $268 per credit hour

% increase= (Final - initial )/initial *100

substitute

% increase= (268- 99 )/268 *100

% increase= 169 /268 *100

% increase= 0.630*100

% increase= 63%

Hence the increase in the tuition from 1990 to 2003 is 63%

5 0
2 years ago
Which ordered pairs in the form (x, y) are solutions to the equation<br><br> 7x−5y=28?
belka [17]

Answer:7x + -5y = -28

(-4, _)

(1, _)

(_, 14)

Step-by-step explanation:

6 0
2 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 <em>more</em> 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 <em>expected</em> 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
Can someone please help me i don’t understand it
forsale [732]

Answer:

d= bv + r

Treat it like you would a normal equation. Multiply by b and add r to isolate d.

5 0
3 years ago
Can someone please help me
horrorfan [7]
Your answer is: Yes, because the sum of the two angles is 180° (the second option).

I hope this helps!
3 0
3 years ago
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