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raketka [301]
3 years ago
8

How much money is $120.00 increased by 70%?

Mathematics
2 answers:
alexdok [17]3 years ago
3 0
The answer to this is 204
swat323 years ago
3 0
70% of 120 is 84 so 
$120 + $84 = $204  

Hope this helps :) 
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Your neighbor needs to put a fence around her front yard and a seperate fence around her backyard. Her front yard is 65 feet lon
elena-14-01-66 [18.8K]
Front yard
65+65+45+45=220
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45+45+35+35=160
220+160=380
380 feet
5 0
3 years ago
Simplify the expression. (3.5x - 5) + (7.2x + 10)
Andru [333]

Answer: 10.7x + 5

Step-by-step explanation:

(3.5 + 7.2)x + (-5 + 10) = 10.7x + 5

8 0
3 years ago
Which of the following is equal to (3^-3)(3^7)?<br><br> a. 3^4<br> b. 3^-4<br> c. 3^-10<br> d. 3^10
balu736 [363]

a. 3^4 which is equal to (3^-3)(3^7)

5 0
3 years ago
Plz help!!!!!!!!!!!!!!!!!1
Wittaler [7]

Answer:

Area = 45 km

Step-by-step explanation:

The area of a parallelogram = base * height

It is given that:

Base = 6 km

Height = 9 km

When these values are substituted into the are equation, the result is:

Area = 6 * 9

Area = 45 km

Hope this helps :)

3 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
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