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3 years ago

(6y^2-7y-3)-(2y^2-9y-7) simplify

1 answer:

5 0

6y^2-7y-3-2y^2+9y+7 (pretend that you're multiplying -1 to 2y^2-9y-7 which is why 9y and 7 are positive now).
Now you're able to simplify your problem, so...

6y^2-2y^2= 4y^2
-7y+9y= 2y
-3+7=4

4y^2+2y+4

Might want to check my math but that's how you do it.

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3 years ago

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3 years ago

In the diagram of PARALLELOGRAM ABCD, segment BE is perpendicular to segment CED, segment DF is perpendicular to BFC and segment

tiny-mole [99]

Answer:

In parallelogram ABCD

FD is perpendicular to BC

BE is perpendicular to CD

Consider triangle BEC and triangle DFC

FC = EC (Given)

Angle BEC = Angle DFC (=90°)

Angle BCE = Angle DCF (common)

Therefore triangle BEC is congruent to triangle DFC (AAS congruency)

DF = BE (CPCT)

Since the altitudes are equal their bases will also be equal

Therefore BC = DC

Therefore BC = DC = AD = AB

Therefore ABCD is a rhombus

Hope this helps!

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3 years ago

A,D,N, and X are collinear points. D is between N and A. NA+AX=NX. Draw a diagram that represents this information. Help!!!

GREYUIT [131]

I hope this helps you out!

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3 years ago

1. A report from the Secretary of Health and Human Services stated that 70% of single-vehicle traffic fatalities that occur at n

Nuetrik [128]

Using the binomial distribution, it is found that there is a 0.7215 = 72.15% probability that between 10 and 15, inclusive, accidents involved drivers who were intoxicated.

For each fatality, there are only two possible outcomes, either it involved an intoxicated driver, or it did not. The probability of a fatality involving an intoxicated driver is independent of any other fatality, which means that the binomial distribution is used to solve this question.

Binomial probability distribution

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$C_{n,x} = \frac{n!}{x!(n-x)!}$

The parameters are:

x is the number of successes.
n is the number of trials.
p is the probability of a success on a single trial.

In this problem:

70% of fatalities involve an intoxicated driver, hence $p = 0.7$ .

A sample of 15 fatalities is taken, hence $n = 15$ .

The probability is:

$P(10 \leq X \leq 15) = P(X = 10) + P(X = 11) + P(X = 12) + P(X = 13) + P(X = 14) + P(X = 15)$

Hence

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 10) = C_{15,10}.(0.7)^{10}.(0.3)^{5} = 0.2061$

$P(X = 11) = C_{15,11}.(0.7)^{11}.(0.3)^{4} = 0.2186$

$P(X = 12) = C_{15,12}.(0.7)^{12}.(0.3)^{3} = 0.1700$

$P(X = 13) = C_{15,13}.(0.7)^{13}.(0.3)^{2} = 0.0916$

$P(X = 14) = C_{15,14}.(0.7)^{14}.(0.3)^{1} = 0.0305$

$P(X = 15) = C_{15,15}.(0.7)^{15}.(0.3)^{0} = 0.0047$

Then:

$P(10 \leq X \leq 15) = P(X = 10) + P(X = 11) + P(X = 12) + P(X = 13) + P(X = 14) + P(X = 15) = 0.2061 + 0.2186 + 0.1700 + 0.0916 + 0.0305 + 0.0047 = 0.7215$

0.7215 = 72.15% probability that between 10 and 15, inclusive, accidents involved drivers who were intoxicated.

A similar problem is given at brainly.com/question/24863377

5 0

3 years ago