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gizmo_the_mogwai [7]
3 years ago
10

Round nearest hundredth​

Mathematics
1 answer:
maks197457 [2]3 years ago
4 0

Answer:.

Step-by-step explanation:nearest hundredth the answer for this question is joe mama

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Which number has more than two factors? <br> A. 19<br> B. 23<br> C. 25<br> D. 29
Whitepunk [10]

The answer would be 25 because all of the others are prime numbers, meaning they can only be divided by themselves and 1. So it would be C.
4 0
3 years ago
What is nth for 4, 8, 16, 32?
daser333 [38]

Answer: nth=4x

Step-by-step explanation: 4,8,16,32

So when x=1 nth=4

Then 4x should be the answer we need to find any term

7 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
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.

6 0
3 years ago
The length of a bridge is 230 m. The length of each of Robert's paces is 75 cm.
wariber [46]

Answer:

307 paces

Step-by-step explanation:

First, lets convert Robert's pace to metres. (cm > m)

75cm = (75/100)m

= 0.75m

Number of Full Paces = Length of Bridge / 1 Full Pace

=

\frac{230m}{0.75m}  \\  = 306 \frac{2}{3} paces

See , we must have full paces, and in this case we are finding the minimum number of paces, so we will be looking at the lowest possible whole number, which in this case, is 307 paces.

5 0
2 years ago
I need a bit of help with this, May you please help me?<br>15 points. ​
max2010maxim [7]

The answer is 7 feet becaise 0 opp ik

7 0
3 years ago
Read 2 more answers
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