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2 years ago

Plssssss help! I'm timed

Mathematics

1 answer:

frozen [14]2 years ago

4 0

Answer:

B: f(x) = –7x2 – x + 2

Step-by-step explanation:

The biggest exponent in a quadratic function is a 2.

A. Is NOT the answer because it has an exponent of 3 (that's a cubic not a quadratic)

C. Is NOT the answer because it is linear, if you graph it, it's a line. The highest exponent is 1.

D. Is NOT the answer, because even though it shows a power of 2, the coefficient (the number next to it) is a zero, so that term would fall out of the equation and it would be linear like C.

So B: f(x) = –7x2 – x + 2 is your answer. Its highest exponent is two and if you graph it, it would make a parabola.

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Can someone help me with part B? Thank you! Brainliest for correct answer!

Ivanshal [37]

Answer: The mean is 5

Step-by-step explanation:

If you add all of the numbers and divide by how much numbers there are (8), then you get 5. Then the mean absolute deviation is the number away from the mean.

So I will put the mean absolute deviation in order of the numbers

4 (1 away from mean), 3 (2 away from the mean), 4( 1 away from the mean), 5 (same number as the mean), 6 (1 away from the mean), 6(1 away from mean), 8 (3 away from mean), and, finally another 4 (1 away)

Hope this helps :)

3 0

3 years ago

Read 2 more answers

Jenn says there are 2 weeks 5 days left in the year. How many days are left in a year

Law Incorporation [45]

19 days are left in a year because 1 week=7 days which means 2 weeks=14 days 14 +5=19

4 0

3 years ago

Read 2 more answers

Examine the division problem

11111nata11111 [884]

Answer:

-9/2 (4/3)

-6

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

An identification code is to consist of 2 letters followed by 3 digits. how many different codes are possible if repetition is p

Vedmedyk [2.9K]

I like the 'question mark' at the end:

26 letters (English), 10 digits (0,1,...9)

26*26*10*10*10 = 676000

Now, this was for LLDDD (Letter Letter Digit Digit Digit)

It seems this is the order they want. Other wise, one should multiply by all the combinations in the order: LDLDD, DLLDD, etc ... (10 combinations), but again it seems they want this LLDDD

So 676000

5 0

3 years ago

Seventy percent of all vehicles examined at a certain emissions inspection station pass the inspection. Assuming that successive

LenaWriter [7]

Answer:

(a) The probability that all the next three vehicles inspected pass the inspection is 0.343.

(b) The probability that at least 1 of the next three vehicles inspected fail is 0.657.

(d) The probability that at most 1 of the next three vehicles passes is 0.216.

(e) The probability that all 3 vehicle passes given that at least 1 vehicle passes is 0.3525.

Step-by-step explanation:

Let X = number of vehicles that pass the inspection.

The probability of the random variable X is P (X) = 0.70.

(a)

Compute the probability that all the next three vehicles inspected pass the inspection as follows:

P (All 3 vehicles pass) = [P (X)]³

$=(0.70)^{3}\\=0.343$

Thus, the probability that all the next three vehicles inspected pass the inspection is 0.343.

(b)

Compute the probability that at least 1 of the next three vehicles inspected fail as follows:

P (At least 1 of 3 fails) = 1 - P (All 3 vehicles pass)

$=1-0.343\\=0.657$

Thus, the probability that at least 1 of the next three vehicles inspected fail is 0.657.

(c)

Compute the probability that exactly 1 of the next three vehicles passes as follows:

P (Exactly one) = P (1st vehicle or 2nd vehicle or 3 vehicle)

= P (Only 1st vehicle passes) + P (Only 2nd vehicle passes)

+ P (Only 3rd vehicle passes)

$=(0.70\times0.30\times0.30) + (0.30\times0.70\times0.30)+(0.30\times0.30\times0.70)\\=0.189$

Thus, the probability that exactly 1 of the next three vehicles passes is 0.189.

(d)

Compute the probability that at most 1 of the next three vehicles passes as follows:

P (At most 1 vehicle passes) = P (Exactly 1 vehicles passes)

+ P (0 vehicles passes)

$=0.189+(0.30\times0.30\times0.30)\\=0.216$

Thus, the probability that at most 1 of the next three vehicles passes is 0.216.

(e)

Let X = all 3 vehicle passes and Y = at least 1 vehicle passes.

Compute the conditional probability that all 3 vehicle passes given that at least 1 vehicle passes as follows:

$P(X|Y)=\frac{P(X\cap Y)}{P(Y)} =\frac{P(X)}{P(Y)} =\frac{(0.70)^{3}}{[1-(0.30)^{3}]} =0.3525$

Thus, the probability that all 3 vehicle passes given that at least 1 vehicle passes is 0.3525.

7 0

3 years ago