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3 years ago

A fair coin is tossed 9 times. what is the probability that + "at most 6 heads" + appear?

Mathematics

1 answer:

kherson [118]3 years ago

7 0

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Which represents the solution set of an any quality 5X-9 greater than 21

Svet_ta [14]

Answer:

The last one

Step-by-step explanation:

5x<30

x<6

Because when you divide by pos+ number

the symbol cannot change.

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2 years ago

Identify the percent of change as an increase or a decrease. 50 pounds to 35 pounds

mojhsa [17]

Answer:

There is a 30% decrese.

Step-by-step explanation:

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2 years ago

HELP PLEASE Combine like terms to create an equivalent expression.

k0ka [10]

SOLUTION:

= ( 7 / 8 )m + 9 / 10 - 2m - 3 / 5

= - ( 9 / 8 )m + 3 / 10

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3 years ago

Mrs. chill has 26 yards of ribbon. She places equal amounts on 3 different spools to sell. How much ribbon in yards are on each

den301095 [7]

8.6 yards are on each spool

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2 years ago

How many times must we toss a coin to ensure that a 0.95-confidence interval for the probability of heads on a single toss has l

musickatia [10]

Answer:

(1) 97

(2) 385

(3) 9604

Step-by-step explanation:

The (1 - α) % confidence interval for population proportion is:

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

The margin of error in this interval is:

$MOE= z_{\alpha/2}\sqrt{\frac{\hat p(1-\hat p)}{n}}$

The formula to compute the sample size is:

$\\n=\frac{z_{\alpha/2}^{2}\times \hat p(1-\hat p)}{MOE^{2}}$

(1)

Given:

$\hat p = 0.50\\MOE=0.1\\z_{\alpha/2}=z_{0.05/2}=z_{0.025}=1.96$

*Use the z-table for the critical value.

Compute the value of n as follows:

$\\n=\frac{z_{\alpha/2}^{2}\times \hat p(1-\hat p)}{MOE^{2}}\\=\frac{1.96^{2}\times0.50\times(1-0.50)}{0.1^{2}}\\=96.04\\\approx97$

Thus, the minimum sample size required is 97.

(2)

Given:

$\hat p = 0.50\\MOE=0.05\\z_{\alpha/2}=z_{0.05/2}=z_{0.025}=1.96$

*Use the z-table for the critical value.

Compute the value of n as follows:

$\\n=\frac{z_{\alpha/2}^{2}\times \hat p(1-\hat p)}{MOE^{2}}\\=\frac{1.96^{2}\times0.50\times(1-0.50)}{0.05^{2}}\\=384.16\\\approx385$

Thus, the minimum sample size required is 385.

(3)

Given:

$\hat p = 0.50\\MOE=0.01\\z_{\alpha/2}=z_{0.05/2}=z_{0.025}=1.96$

*Use the z-table for the critical value.

Compute the value of n as follows:

$\\n=\frac{z_{\alpha/2}^{2}\times \hat p(1-\hat p)}{MOE^{2}}\\=\frac{1.96^{2}\times0.50\times(1-0.50)}{0.01^{2}}\\=9604$

Thus, the minimum sample size required is 9604.

8 0

2 years ago