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liberstina [14]
2 years ago
8

Try to determine which set of digits ends product three consecutive natural numbers?​

Mathematics
1 answer:
kirill115 [55]2 years ago
8 0

Answer:

0, 4, 6

Step-by-step explanation:

1 * 2 * 3 = 6

2 * 3 * 4 = 24

3 * 4 * 5 = 60

4 * 5 * 6 = 120

5 * 6 * 7 = 210

6 * 7 * 8 = 336

7 * 8 * 9 = 504

etc.

Answer: 0, 4, 6

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If after 45 minutes han cycled for 6 miles and claire cycled for 9 miles what will the difference between them be after 45 minut
IceJOKER [234]

Answer:

The difference between them will be 6.

Step-by-step explanation:

When they went for their first round of cycling, which was 45 minutes, it was already a distance of 3 miles adding another 6 to Han's miles and another 9 to Claire's miles it would give you 18 and 12 and 18 minus 12 is 6.

Therefore, it was doubled from the distance only being 3 miles to 6 miles.

5 0
2 years ago
eric is making a dress. the design calls for a right triangular piece of material. if the second angle has to be 35% what is the
svet-max [94.6K]
U have a right triangle...so one angle is 90%......and the second angle is 35%...keep in mind, the angles of a triangle = 180%
so the measure of the third angle is : 180 - 90 - 35 = 55%
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2 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
Gala2k [10]

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}

(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:

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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Using the segment addition postulate, which is true?
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