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Bas_tet [7]
3 years ago
6

Explain how an estimate helps you to place the decimal point when mutiplyning 3.9 times 5.3

Mathematics
2 answers:
kvv77 [185]3 years ago
7 0
It'll make it easier to multiply. If you round 3.9 to four, and 5.3 to five, it's quite easy to figure out the answer is about 20. I hope this helps! :)
Juli2301 [7.4K]3 years ago
6 0
You can round these numbers to the closes whole number; 3.9-->4 and 5.3-->5, and then you can do 5x4 to get 20; and if you multiply regularly without estimating, your answer will be around 20.
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On a certain hot summer day, 481 people used the public swimming pool. The daily prices are 1.25 for children and 2.25 for adult
spayn [35]

Answer:the number of children that swam in the public pool that day is 222

the number of adult that swam in the public pool that day is 259

Step-by-step explanation:

Let x represent the number of children that swam in the public pool that day.

Let y represent the number of adult that swam in the public pool that day.

On a certain hot summer day, 481 people used the public swimming pool. This means that

x + y = 481

The daily prices are 1.25 for children and 2.25 for adults. The receipts for admission totaled to 86.25. This means that

1.25x + 2.25y = 860.25 - - - - - - - -1

Substituting x = 481 - y into equation 1, it becomes

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x = 481 - 259

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Step-by-step explanation:

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Gala2k [10]

Answer:

(a) There are 70 different ways set up 4 computers out of 8.

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Step-by-step explanation:

Of the 9 new computers 4 are laptops and 5 are desktop.

Let X = a laptop is selected and Y = a desktop is selected.

The probability of selecting a laptop is = P(Laptop) = p_{X} = \frac{4}{9}

The probability of selecting a desktop is = P(Desktop) = p_{Y} = \frac{5}{9}

Then both X and Y follows Binomial distribution.

X\sim Bin(9, \frac{4}{9})\\ Y\sim Bin(9, \frac{5}{9})

The probability function of a binomial distribution is:

P(U=k)={n\choose k}\times(p)^{k}\times (1-p)^{n-k}

(a)

Combination is used to determine the number of ways to select <em>k</em> objects from <em>n</em> distinct objects without replacement.

It is denotes as: {n\choose k}=\frac{n!}{k!(n-k)!}

In this case 4 computers are to selected of 8 to be set up. Since there cannot be replacement, i.e. we cannot set up one computer twice or thrice, use combinations to determine the number of ways to set up 4 computers of 8.

The number of ways to set up 4 computers of 8 is:

{8\choose 4}=\frac{8!}{4!(8-4)!}\\=\frac{8!}{4!\times 4!} \\=70

Thus, there are 70 different ways set up 4 computers out of 8.

(b)

It is provided that 4 computers are randomly selected.

Compute the probability that exactly 3 of the 4 computers selected are desktops as follows:

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Thus, the probability that exactly three of the selected computers are desktops is 0.305.

(c)

Compute the probability that of the 4 computers selected at least 3 are desktops as follows:

P(Y\geq 3)=1-P(Y

Thus, the probability that at least three of the selected computers are desktops is 0.401.

6 0
3 years ago
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