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3 years ago

Need help on this question

Mathematics

2 answers:

klasskru [66]3 years ago

4 0

Answer:

Step-by-step explanation:

Kay [80]3 years ago

3 0

Answer:

option b

i.e-

(-1,3) and (2,6)

Step-by-step explanation:

The solution of two graphs are the point where they intersect each other.

Above, they intersect each other at two points

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Can i get the answer

Leokris [45]

Answer:

2.6

Step-by-step explanation:

by similar property

$HJ=\frac{JI\times EG}{GF}\\x+16=\frac{31\times 36}{60}\\x+16=18.6\\x=2.6$

5 0

3 years ago

ACT mathematics score for a particular year are normally distributed with a mean of 27 and a standard deviation of 2 points.

mel-nik [20]

Answer:

A: 16%

B: 95%

C: 97.5%

Step-by-step explanation:

According to the empirical rule:

68% of a normal distribution is between -1 and +1 standard deviations.

95% of a normal distribution is between -2 and +2 standard deviations.

99.7% of a normal distribution is between -3 and +3 standard deviations.

Given μ = 27 and σ = 2.

Part A

29 is one standard deviation above the mean. We can show this by calculating the z-score:

z = (x − μ) / σ

z = (29 − 27) / 2

z = 1

We know that 68% is between -1 and +1 standard deviations. Since normal distributions are symmetrical, we can also say that 34% is between 0 and +1 standard deviations.

P(0 < Z < 1) = 68%/2 = 34%

We can also say that 50% is less than 0 standard deviations.

P(Z < 0) = 50%

Therefore, P(Z < 1) = 34% + 50% = 84%.

Which means P(Z > 1) = 100% − 84% = 16%.

Part B

Like before, calculate the z-scores:

z₁ = (23 − 27) / 2

z₁ = -2

z₂ = (31 − 27) / 2

z₂ = 2

From the empirical rule, we know this is 95% of the normal distribution.

Part C

We found in part B that the z-score is 2.

P(0 < Z < 2) = 95%/2 = 47.5%

P(Z < 2) = 50% + 47.5% = 97.5%

5 0

4 years ago

Please help!! I need help with this question.

Ainat [17]

Its kinda blurry and I can't read the whole question

4 0

4 years ago

Could you please look at this question?

MrMuchimi

Answer:

a. see attached

b. H(t) = 12 -10cos(πt/10)

c. H(16) ≈ 8.91 m

Step-by-step explanation:

The cosine function has its extreme (positive) value when its argument is 0, so we like to use that function for circular motion problems that have an extreme value at t=0. The midline of the function needs to be adjusted upward from 0 to a value that is 2 m more than the 10 m radius. The amplitude of the function will be the 10 m radius. The period of the function is 20 seconds, so the cosine function will be scaled so that one full period is completed at t=20. That is, the argument of the cosine will be 2π(t/20) = πt/10.

The function describing the height will be ...

H(t) = 12 -10cos(πt/10)

The graph of it is attached.

See part a.

The wheel will reach the top of its travel after 1/2 of its period, or t=10. Then 6 seconds later is t=16.

H(16) = 12 -10cos(π(16/10) = 12 -10cos(1.6π) ≈ 12 -10(0.309017) ≈ 8.90983

The height of the rider 6 seconds after passing the top will be about 8.91 m.

8 0

3 years ago

242.8333333333333 rounded to the nearest hundredth

balandron [24]

The answer would be 242.8
It will not go up because in order to round up it must be 5 or above.

8 0

4 years ago

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