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2 years ago

A fair coin is flipped 120 times. Estimate the expected number of ‘heads’.

Mathematics

2 answers:

Bumek [7]2 years ago

7 0

when coin is flipped 5 times probability is 1/2

so coin heads comes out to be 2.5

This is a binomial distribution with a probability of success = 0.5 (fair coin).

n=10

mean = E[x] = np =10*0.5 =5

Naily [24]2 years ago

6 0

Answer:

76.47

How I did it(using a specific range)

(n, 54, 66 (120 C)) 2¹²⁰ = 0.76 × 100 ≈ 76.47

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The answer is 4 2/3 minus 2

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2 years ago

Camila wanted to create a second garden to grow even more vegetables. To determine the dimensions of her second garden, she used

OlgaM077 [116]

Using proportions, considering original measures of 10 feet, we have that:

The horizontal measurement of the second garden is of 7.5 feet.

The vertical measurement of the second garden is of 15 feet.

<h3>What is a proportion?</h3>

A proportion is a fraction of total amount.

In this problem, the total amount is the original measures of 10 feet.

The horizontal measurement has a scale factor of 75%, hence:

$h = 0.75(10) = 7.5$

The vertical measurement has a scale factor of 150%, hence:

$h = 1.5(10) = 15$

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2 years ago

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11Alexandr11 [23.1K]

Answer:

11,5 cm or 23 squares

Step-by-step explanation:

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2 years ago

The town of Hayward (CA) has about 50,000 registered voters. A political research firm takes a simple random sample of 500 of th

timama [110]

Answer: (0.076, 0.140)

Step-by-step explanation:

Confidence interval for population proportion (p) is given by :-

$\hat{p}\pm z_{\alpha/2}\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}$

, where $\hat{p}$ = sample proportion.

n= sample size.

$\alpha$ = significance level .

$z_{\alpha/2}$ = critical z-value (Two tailed)

As per given , we have

sample size : n= 500

The number of Independents.: x= 54

Sample proportion of Independents $\hat{p}=\dfrac{x}{n}=\dfrac{54}{500}=0.108$

Significance level 98% confidence level : $\alpha=1-0.98=0.02$

By using z-table , Critical value : $z_{\alpha/2}=z_{0.01}=2.33$

The 98% confidence interval for the true percentage of Independents among Haywards 50,000 registered voters will be :-

$0.108\pm (2.33)\sqrt{\dfrac{0.108(1-0.108)}{500}}\\\\=0.108\pm2.33\times0.013880634\\\\=0.108\pm0.03234187722\\\\\approx0.108\pm0.032=(0.108-0.032,\ 0.108+0.032)=(0.076,\ 0.140)$

Hence, the 98% confidence interval for the true percentage of Independents among Haywards 50,000 registered voters.= (0.076, 0.140)

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2 years ago

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shepuryov [24]

Answer:

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Simplify by adding terms.

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2 years ago