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barxatty [35]
2 years ago
6

How i can answer this question?, NO LINKS, if you answer correctly i'll give u brainliest!

Mathematics
2 answers:
Jobisdone [24]2 years ago
8 0

6(4a-3b)

if you draw a factor tree that goes like:

    24

1         24

2        12

...       ...  (and so on, you'll find 6 is the highest)

slavikrds [6]2 years ago
5 0
6•(4a-3b)

Steps:

Rewrite -18 as 3• 6
Rewrite 24 as 4 • 6
= 4 • 6a + 3• 6b

= 6(4a-3b)
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A square garden has sides 100 ft long. You want to build a brick path along a diagonal of the square. How long will the path be?
denis-greek [22]
100^2+100^2=c^2
10,000+10,000=c^2
20,000=c^2
141=c
7 0
3 years ago
The Klein family is taken a weekend trip to their grandparents who live 134 miles away. Mr. Klein always drives at a
kogti [31]

Answer:

92 more miles

Step-by-step explanation:

60 mph is 1 mile per min so in 42 min they traveled 42 miles and 134 miles minus 42 miles is 92 miles

8 0
3 years ago
What is the value of x in the equation 5/6x + (-1 1/3) = -1
irga5000 [103]

Solve for  x  by simplifying both sides of the equation, then isolating the variable.

Exact Form: x= 2/5

Decimal Form: x= 0.4

5 0
3 years ago
Suppose the number of insect fragments in a chocolate bar follows a Poisson process with the expected number of fragments in a 2
leonid [27]

Answer:

a)The expected number of insect fragments in 1/4 of a 200-gram chocolate bar is 2.55

b)0.6004

c)19.607

Step-by-step explanation:

Let X denotes the number of fragments in 200 gm chocolate bar with expected number of fragments 10.2

X ~ Poisson(A) where \lambda = \frac{10.2}{200} = 0.051

a)We are supposed to find the expected number of insect fragments in 1/4 of a 200-gram chocolate bar

\frac{1}{4} \times 200 = 50

50 grams of bar contains expected fragments = \lambda x = 0.051 \times 50=2.55

So, the expected number of insect fragments in 1/4 of a 200-gram chocolate bar is 2.55

b) Now we are supposed to find the probability that you have to eat more than 10 grams of chocolate bar before ending your first fragment

Let X denotes the number of grams to be eaten before another fragment is detected.

P(X>10)= e^{-\lambda \times x}= e^{-0.051 \times 10}= e^{-0.51}=0.6004

c)The expected number of grams to be eaten before encountering the first fragments :

E(X)=\frac{1}{\lambda}=\frac{1}{0.051}=19.607 grams

7 0
3 years ago
Free poinnttssssssssssssssss and yourr welcomeee
DerKrebs [107]

Answer:

thank youuuuuu

Step-by-step explanation:

have a good day

5 0
3 years ago
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