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GrogVix [38]
3 years ago
15

What is the scale factor for this?

Mathematics
1 answer:
Alenkinab [10]3 years ago
3 0
SF = 5/2 = 2.5

answer is 5/2 or 2.5
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How do u simply 2x=22<br> Confused about this one .
abruzzese [7]

44     22 + 22 equals 44

         

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
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
2 years ago
This jar has 1 red, 74 yellow, and 25 green gumballs. What are the chances of picking a green gumball? the red one? ​
kari74 [83]

Answer: green = 25/100, red = 1/100

Step-by-step explanation: add ALL of the gumballs, there's your denominator, now you have 0/100 (1+74+25), count the amount of marbles you want to get the probability of, lets say green, there's your numerator (25) = 25/100

5 0
3 years ago
Which expression has the same value as -4(5 - x)?
grigory [225]
None of these is correct.
-4(5-x)
= (-4•5) + (-4•-x)
= -20 + 4x
= 4x-20
4 0
2 years ago
Read 2 more answers
Make a conjecture about the sum of the first 15 positive even number
erica [24]
I conclude that the sum will be even because any even number can be represented by 2n where n is a whole number
and even numbers are 2 apart, so
the sum of the first 15 are
2n+2(n+1)+2(n+2) etc until we get to 2(n+14)
we can undistribute the 2 from all of them and get
2(n+n+1+n+2...n+14)
and we are sure that whatever is in the parenthasees is a whole number because whole+whole=whole
therefor, the sum is even

if you did want to find the sum then
an=2n
the 15th even number is 30
the first is 2
S15=15(2+30)/2=15(32)/2=15(16)=240
which is even
6 0
3 years ago
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