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almond37 [142]
3 years ago
14

Evaluate: 8 – 4(5) + 12. Work and explanation please!

Mathematics
2 answers:
goldfiish [28.3K]3 years ago
8 0
8-4(5)+12
8-20+12
20-20
0

Hope this helps :)
gogolik [260]3 years ago
3 0
8-4(5)+12
8-20+12
-12+12
0
That should be your answer
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Help me with this pls
Rufina [12.5K]

Answer:

b^-6

Step-by-step explanation:

(b^-2) / (b^4) = b^(-2 - 4)

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Which lines are parallel? Check all that appley
7nadin3 [17]

Answer:

Do you have a diagram? I cant help you if you have a diagram.

Step-by-step explanation:

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3 years ago
You are offered a 30--day trial period at your dream job. However, the owner of the company is a little unusual so the pay optio
S_A_V [24]

Answer:

The question is not complete but from the flow of thought, the question seems to be about what your choice will be between option 1 and option 2.

The correct answer is

option 2

Step-by-step explanation:

The question is asking you to choose an option that pays more after 30 days. In other to determine the more beneficial option, we will compare both options and select the one that pays more. This is done as follows:

option 1: $60,000 per day, after 30 days =

60,000 × 30 = $1,800,000

option 2 = $5,368,709.12 after 30 days

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4 0
3 years ago
What percentage of people has an IQ score less than 83 or greater than 117
Julli [10]

Answer:

(a) 95% of people have an IQ score between 66 and 134.

(b) 32% of people have an IQ score of less than 83 or greater than 117​.

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Step-by-step explanation:

7 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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