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iris [78.8K]
2 years ago
8

75 POINTS

Mathematics
1 answer:
Gnoma [55]2 years ago
5 0

Answer:

a) 235.62 cm³ (nearest hundredth)

b) 2. tripling the height to 9 cm, while the radius remains 5 cm, gives the most volume

Step-by-step explanation:

volume of a cylinder = 2\pi r^2h  (where r is the radius and h is the height)

Given:

  • r = 5 cm
  • h = 3 cm

a) ⇒ volume = \pi \times 5^2 \times 3=75\pi =235.62 cm³ (nearest hundredth)

b)  Given:

  • r = 10 cm
  • h = 3 cm

⇒ volume = \pi \times 10^2 \times 3=300\pi =942.48 cm³ (nearest hundredth)

Given:

  • r = 5 cm
  • h = 9 cm

⇒ volume = \pi \times 5^2 \times 9=225\pi =706.86 cm³ (nearest hundredth)

Therefore, 2) tripling the height to 9 cm, while the radius remains 5 cm, gives the most volume

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Answer:

When V is composed with r, the composite function is StartFraction 1 Over 48 EndFraction pi t Superscript 6

V(r(6)) shows that the volume is 972π inches cubed when the radius increases to 6 inches.

Step-by-step explanation:

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Solve for v simplify your answer as much as possible: -2 + 3v = -23​
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Step-by-step explanation:

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Answer:

a) The study is an observational study

b) The proportion of children with significant social activity in children with acute lymphoblastic leukemia is approximately 0.802

The proportion of children with significant social activity in children without acute lymphoblastic leukemia is approximately 0.8565

c) The odds ratio is approximately 0.6780

d) The 95% CI is approximately 0.523 < OR < 0.833

e) i) Yes

ii) Children with more social activity have a higher occurrence of acute lymphoblastic leukemia

Step-by-step explanation:

a) An experimental study is a study in which the researcher adds inputs to the study and then monitors the outcomes

An observational study is one in which the researcher observes risk factors and does not intervene in the process

Therefore, the study is an observational study

b) The proportion of children with significant social activity in children with acute lymphoblastic leukemia is \hat p_1 = 1020/1272 = 85/106 = 0.8\overline {0188679245283} ≈ 0.802

The proportion of children with significant social activity in children without acute lymphoblastic leukemia is \hat p_2 = 5343/6238 ≈ 0.8565

c) We have the following two way table;

\begin{array}{ccc}Lymphoblastic \ Leukemia &With \ Social \ Activity & Without \ Social \ Activity\\With&1020 \ (a)&252 \ (b)\\Without&5343 \ (c)&895 \ (d) \end{array}

The \ odds \ ratio = \dfrac{1,020 \times 895}{5,343 \times 252} \approx 0.6780

The odds ratio ≈ 0.6780

d) The 95% confidence interval for the odds ratio is given as follows;

95 \% \ CI = OR \ \pm \ 1.96 \times \sqrt{\dfrac{1}{a} +\dfrac{1}{b}+\dfrac{1}{c}+\dfrac{1}{d}}

Where;

a = 1020, b = 252, c = 5343, d = 895

Therefore;

95 \% \ CI \approx  0.6780 \ \pm \ 1.96 \times \sqrt{\dfrac{1}{1,020} +\dfrac{1}{252}+\dfrac{1}{5,343}+\dfrac{1}{895}} \approx 0.678 \pm0.155

The 95% CI ≈ 0.523 < OR < 0.833

e) i) Given that the observed Odds Ratio is within the Confidence Interval, therefore, there is an indication that the amount of social activity is associated with acute lymphoblastic leukemia

ii) Based on the proportion of the study findings, children with more social activity have a higher occurrence of acute lymphoblastic leukemia

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Answer:

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Find the sum of the interior angles of a 15 gon
MA_775_DIABLO [31]
The sum of that is the answer = 3
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