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3 years ago

Read this and answer this for me Thanks <3

Mathematics

1 answer:

Norma-Jean [14]3 years ago

6 0

Answer:

The answer is 1st point

Step-by-step explanation:

(12x²+4x-5)-(9x²-9x+1)

= 12x²+4x-5-9x²+9x-1

= 3x²+13x-6

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Jacob needs less than 5 C's on his transcript to qualify for UGA. He already has 1 C. At most, how many more C's can he get and

victus00 [196]

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4 years ago

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Which linear system of equations does the matrix represent 4 -7 -14 4 -12 4

Novosadov [1.4K]

The given matrix represent the following linear system

4x - 7y = -12
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3 years ago

A class has 12 boys and 4 girls. What is the ratio of boys to girls written as a decimal?

ss7ja [257]

Answer:

C: 3.0

Step-by-step explanation:

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3 years ago

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how is building a triangle with different side lengths given different than building a quadrilateral with given side lengths?

Mariana [72]

Answer:

there can only be one possibility for a triangle when given the lengths of all the sides but for a quadrilateral the measure of the angles could differ depending on the person building the,. this is because triangles are more stable than quadrilaterals meaning that their side lengths follow a lot more rules than quadrilaterals do, for example the length of the side lengths can indicate whether or not that triangle is an acute, obtuse, or right triangle, and this is also evident by considering that you can use the SSS theorem to indicate two triangles are congruent, but for quadrilaterals you cant do that

Step-by-step explanation:

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4 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 :

$E(X)=\frac{1}{\lambda}=\frac{1}{0.051}=19.607 gram$ s

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3 years ago