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2 years ago

Given the polynomial function below, find F(2).

Mathematics

1 answer:

olganol [36]2 years ago

3 0

Answer:

C) 35

Step-by-step explanation:

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Plz plz help im not good at math..................................at all

asambeis [7]

Answer:

8: subtract 1.50 from 11.50 until you get to 7.00 and count how many times you subtracted it that is your answer

9: take 254 and subtract 4 until you get to 228 count how many times you subtracted and that is your answer

10: take ten and add 6 until you get to 46 count how many times you added and that is your answer

Step-by-step explanation:

6 0

2 years ago

Is this the correct answer

enot [183]

Answer:

It should be correct.

8 0

2 years ago

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Yuliya22 [10]

Step-by-step explanation:

1+2√2 ²π = 9+4√2 π cm² thats my answer

6 0

2 years ago

During a recent drought, a water utility in a certain town sampled 100 residential water bills and found out that 73 of the resi

ankoles [38]

Answer:

(0.6430, 0.8170)

Step-by-step explanation:

Given that during a recent drought a water utility in a certain town sampled 100 residential water bills and found out that 73 of the residences had reduced their water consumption over that of the previous year.

Sample size n = 100

Sample proportion p = $\frac{73}{100} =0.73$

q = 1-p = 0.23

Std error of proportion = $\sqrt{\frac{pq}{n} } \\=\sqrt{\frac{0.73*0.27}{100} } \\=0.04440$

95% Z critical value = 1.96

Margin of error = $1.96*0.0444 = 0.0870$

Confidence interval = sample proportion ±margin of error

0.642983946

0.817016054

(0.6430, 0.8170)

4 0

2 years ago

I need help with this problem! Thank you!

QveST [7]

Answer:

The amount of water in the reservoir at the beginning of second week

$w_{2} = 299.8800$

Step-by-step explanation:

Step(i):-

Given equation

w₀ = 300 and

$w_{n} = (0.9993) w _{n-1} + 0.15$

Put n =1

$w_{1} = (0.9993) w _{1-1} + 0.15$

$w_{1} = (0.9993) w _{0} + 0.15$

$w_{1} = (0.9993) X 300 + 0.15$

$w_{1} = 299.94$

Step(ii):-

Put n =2

$w_{2} = (0.9993) w _{2-1} + 0.15$

$w_{2} = (0.9993) w _{1} + 0.15$

$w_{2} = (0.9993) X (299.94) + 0.15$

$w_{2} = 299.8800$

Conclusion:-

The amount of water in the reservoir at the beginning of second week

$w_{2} = 299.8800$

3 0

2 years ago