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Step2247 [10]
3 years ago
9

What is the factorization of 3x2 – 8x + 5?

Mathematics
1 answer:
tatuchka [14]3 years ago
5 0

For this case we must factor the following expression:

3x ^ 2-8x + 5

We rewrite the middle term as a sum of two terms whose product is

a * c = 3 * 5 = 15

And whose sum is

b = -8

These numbers are:

-5 and -3

So:

3x ^ 2-5x-3x + 5

We factor the maximum common denominator of each group:x (3x-5) - (3x-5)

We take common factor(3x-5):

(3x-5) (x-1)

Answer:

(3x-5) (x-1)

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A. You bring $40 to the carnival. You need to keep at least $5 for the bus ride home. how many $3 dollar rides can you go on

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Yahoo creates a test to classify emails as spam or not spam based on the contained words. This test accurately identifies spam (
Amiraneli [1.4K]

Answer and Step-by-step explanation:

The computation is shown below:

Let us assume that

Spam Email be S

And, test spam positive be T

Given that

P(S) = 0.3

P(\frac{T}{S}) = 0.95

P(\frac{T}{S^c}) = 0.05

Now based on the above information, the probabilities are as follows

i. P(Spam Email) is

= P(S)

= 0.3

P(S^c) =  1 - P(S)

= 1 - 0.3

= 0.7

ii. P(\frac{S}{T}) = \frac{P(S\cap\ T}{P(T)}

= \frac{P(\frac{T}{S}) . P(S) }{P(\frac{T}{S}) . P(S) + P(\frac{T}{S^c}) . P(S^c) }

= \frac{0.95 \times 0.3}{0.95 \times 0.3 + 0.05 \times 0.7}

= 0.8906

iii. P(\frac{S}{T^c}) = \frac{P(S\cap\ T^c}{P(T^c)}

= \frac{P(\frac{T^c}{S}) . P(S) }{P(\frac{T^c}{S}) . P(S) + P(\frac{T^c}{S^c}) . P(S^c) }

= \frac{(1 - 0.95)\times 0.3}{ (1 -0.95)0.95 \times 0.3 + (1 - 0.05) \times 0.7}

= 0.0221

We simply applied the above formulas so that the each part could come

8 0
2 years ago
Choose one of the factors of 6x^3 + 6
ivanzaharov [21]

Answer: First option is correct.

Step-by-step explanation:

Since we have given that

6x^3+6

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6x^3+6\\=6(x^3+1)

Now, we use the formula i.e.

a^3+b^3=(a+b)(a^2-ab+b^2)

By using this, we get ,

6(x^3+1)\\=6(x+1)(x^2-x+1)

So,

(x+1) \text{ is the factor of }6x^3+6.


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Given sinθ=- 3/5 and cscθ=-5/3 in quadrant III, find the value of other trigonometric functions using a Pythagorean Identity. Sh
Yakvenalex [24]
Let ABC be a triangle in the 3rd quadrant, right-angled at B. 
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 ă€–BC〗^2=16

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 cotθ=Base/Perpendicular cotθ=4/3
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