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Veseljchak [2.6K]
3 years ago
14

What is the remainder when 849,400 is divided by 216​

Mathematics
2 answers:
S_A_V [24]3 years ago
5 0
The anserws is 88 so your welcome
Mnenie [13.5K]3 years ago
5 0

Answer:

<em>The remainder is 88</em>

Step-by-step explanation:

<u>The Remainder of a Division</u>

We have to find the remainder when we divide 849,400 by 216.

It could be found by doing the long division, but we'll do it with a calculator.

Entering the division in the calculator we get:

849,400/216=3,932.4074...

We take the whole part of the quotient (3,932) and multiply it by the divisor:

3,932*216=849,312

The remainder is obtained by subtracting 849,400-849,312=88

The remainder is 88

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3 0
3 years ago
The computers of nine engineers at a certain company are to be replaced. Four of the engineers have selected laptops and the oth
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Answer:

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

(b) The probability that exactly three of the selected computers are desktops is 0.305.

(c) The probability that at least three of the selected computers are desktops is 0.401.

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}

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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:

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It is provided that 4 computers are randomly selected.

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P(Y=3)={4\choose 3}\times(\frac{5}{9})^{3}\times (1-\frac{5}{9})^{4-3}\\=4\times\frac{125}{729}\times\frac{4}{9}\\  =0.304832\\\approx0.305

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.

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