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satela [25.4K]
3 years ago
12

A web site was hit 300 times over a period of 15 days Show that over some period of 3 consecutive days, it was hit at least 60 t

imes.
Mathematics
1 answer:
marshall27 [118]3 years ago
6 0

Answer:

We will divide the 15 days in five periods of 3 consecutive days each.

Now to solve this we will use the pigeonhole principle.

This states that if (N+1) pigeons occupy N holes, then some hole must have at least 2 pigeons.

So, we have n=300 pigeons  and k=5 holes.

[\frac{n}{k} ]=[\frac{300}{5} ]

Hence, there is a period of 3 consecutive days in which the website was hit at least 60 times.

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Say that you own property with a market price of $74,000. The state tax assessors have given it an assessed value of 48% of that
BigorU [14]

Answer:

A

Step-by-step explanation:

74,000 x .48 = 35520. 35520 is the amount of property you have to pay for.

35520 x 0.031 = 1,101.12.  1.101.12 is the amount you owe in property tax.

5 0
3 years ago
What is the diameter of this circle? A) 2 cm B) 3.14 cm C) 4 cm D) 8 cm
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Answer:

The answer is c

Step-by-step explanation:

Well, since the Circle is 8cm you can divide the circle by two to get 4

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Danni thinks of a number. She subtracts 2 then multiples the result by 8. The answer is the same as subtracting 5 from the numbe
Elena L [17]

Answer:

8

Step-by-step explanation:

let the number=x

8(x-2)=16(x-5)

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8 0
1 year ago
Problem PageQuestion Calcium levels in people are normally distributed with a mean of mg/dL and a standard deviation of mg/dL. I
SIZIF [17.4K]

Complete Question

The complete question is shown on the first uploaded image

Answer:

The  calcium level that is the borderline between low calcium levels and those not considered low is  c = 8.68

Step-by-step explanation:

From the question we are told that

    The mean is  \mu =  9.5\  mg/dL

     The standard deviation \sigma =  0.5 \ mg/dL

    The proportion of the population with low calcium level is  p =5% = 0.05

Let X be a X random calcium level

  Now the  P(X < c) = 0.05

Here P denotes probability

        c is population with calcium level at the borderline

  Since the calcium level is normally distributed the z-value is  evaluated as

    P(Z < \frac{c - \mu}{\sigma } ) = 0.05

The critical value for 0.05 from the standard normal distribution table is

       t_{0.05} = -1.645

=>    \frac{c - \mu}{\sigma }  =  -1.645

substituting values

       \frac{c - 9.5}{0.5 }  =  -1.645

=>    c - 9.5 =  -0.8225

=>     c = 8.68

 

6 0
3 years ago
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