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alukav5142 [94]
3 years ago
6

Which is greater 7500,000 m or 750 km

Mathematics
1 answer:
marta [7]3 years ago
5 0

Step-by-step explanation:

both are equal

pls Mark brainliest

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Solve using substitution: 2x−3y=−1 y=x−1
tekilochka [14]

Answer:

<h2>x = 4</h2><h2>y = 3</h2>

Step-by-step explanation:

2x - 3y = - 1 ………… Equation 1

y = x - 1 ….……….… Equation 2

To solve using substitution , substitute equation 2 into equation 1

That's

2x - 3( x - 1) = - 1

2x - 3x + 3 = - 1

- x = - 1 - 3

- x = - 4

Multiply through by - 1

We have

<h3>x = 4</h3>

Substitute x = 4 , into y = x - 1

We have

y = 4 - 1

y = 3

We have the answers as

x = 4

y = 3

Hope this helps you

6 0
3 years ago
I need help with #9.
Nina [5.8K]
I believe the answer would be 29x + 9 yd. If you add up all the sides with x, it adds up to 29. Then, since perimeter is adding, you add the 9 yds onto that.
7 0
3 years ago
For a party of one hundred fifty people, a caterer will need 67.5 pounds of prime rib. How many packages of prime rib should the
gtnhenbr [62]

Answer:

27 packages

Step-by-step explanation:

67.5/2.5= 27 packages

4 0
3 years ago
Suppose that each coupon obtained is, independently of what has been previously obtained, equally likely to be any of m differen
Triss [41]

ANSWER:

E[X] ≈ m ln m

STEP-BY-STEP EXPLANATION:

Hint: Let X be the number needed. It is useful to represent X by

       m      

X =  ∑  Xi

      i=1

where each Xi  is a geometric random variable

Solution: Assume that there is a sufficiently large number of coupons such that removing a finite number of them does not change the probability that a coupon of a given type is draw. Let X be the number of coupons picked

       m      

X =  ∑  Xi

      i=1

where Xi is the number of coupons picked between drawing the (i − 1)th coupon type and drawing i th coupon type. It should be clear that X1 = 1. Also, for each i:

Xi ∼ geometric \frac{m - i + 1}{m} P r{Xi = n} =(\frac{i-1}{m}) ^{n-1} \frac{m - i + 1}{m}

Such a random variable has expectation:

E [Xi ] =\frac{1}{\frac{m- i + 1}{m}  } = \frac{m}{m-i + 1}

Next we use the fact that the expectation of a sum is the sum of the expectation, thus:

                m           m             m                    m

E[X] = E    ∑  Xi  =   ∑ E   Xi  = ∑  \frac{m}{m-i + 1}  = m ∑ \frac{1}{i} = mHm

               i=1           i=1             i=1                   i=1

In the case of large m this takes on the limit:

E[X] ≈ m ln m

4 0
3 years ago
Can someone help me on any of the problems that are circled
aniked [119]
15. Is $52.50 I think

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3 years ago
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