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3 years ago

What is the probability of getting an odd number on a fair die and a tail on coin flip?

2 answers:

8 0

So we know that the dice had 6 numbers which is 1 to 6, and we know that the odd number between 1 and 6 is 1,3,5. As a result, the probability of rolling an odd number on a fair die is 3/6 or 1/2, and we already know that probability of flipping a tail on a coin is 1/2, so I just take 1/2 times 1/2 to get 1/4 which is C. Hope it help!

Whitepunk [10]3 years ago

3 0

The answer to this question is C)

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2. Find m∠F. a.30° b.60° c.15° d.10°

Juliette [100K]

Answer:

Step-by-step explanation:

$\sf Sin \ F = \dfrac{Opposite \ of \ angle \ F}{hypotenuse}$

$= \dfrac{9}{18}\\\\= 0.5$

$F = Sin^{-1} \ (0.5)\\$

= 30°

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2 years ago

The sum of three numbers is 121 The third number is 3 times the second. The first number is 6 times the second. The first number

Mama L [17]

Answer:

The first number is

72.6

Step-by-step explanation:

Let second number be x.

First number= 6x

Third number= 3x

According to question,

6x+x+3x=121

or, 10x=121

or,x=121/10=12.1

First number= 6x=6×12.1=72.6

Third number=3x=3×12.1=36.3

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2 years ago

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In a survey, the planning value for the population proportion is p* = 0.35. How large a sample should be taken to provide a 95%

lions [1.4K]

Answer:

$n=\frac{0.35(1-0.35)}{(\frac{0.05}{1.96})^2}=349.59$

And rounded up we have that n=350

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

$p$ represent the real population proportion of interest

$\aht p$ represent the estimated proportion for the sample

n is the sample size required (variable of interest)

$z$ represent the critical value for the margin of error

Solution to the problem

The population proportion have the following distribution

$p \sim N(p,\sqrt{\frac{\hat p(1-\hat p)}{n}})$

In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 95% of confidence, our significance level would be given by $\alpha=1-0.95=0.05$ and $\alpha/2 =0.025$ . And the critical value would be given by:

$z_{\alpha/2}=-1.96, z_{1-\alpha/2}=1.96$

The margin of error for the proportion interval is given by this formula:

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

And on this case we have that $ME =\pm 0.05$ and we are interested in order to find the value of n, if we solve n from equation (a) we got:

$n=\frac{\hat p (1-\hat p)}{(\frac{ME}{z})^2}$ (b)

And replacing into equation (b) the values from part a we got:

$n=\frac{0.35(1-0.35)}{(\frac{0.05}{1.96})^2}=349.59$

And rounded up we have that n=350

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3 years ago

Marco thinks the total number of runs a team scores in a season is a better indicator of how many wins it will have in that seas

FrozenT [24]

This is a fair assumption, but not true in every instance. Since home runs do not indicate how many runs score, which are the decider in a win or a loss, they are less accurate at predicting.

5 0

3 years ago

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for a field trip 4 students rode in cars and the rest filled nine buses. how many students were in each bus if 472 students were

Lemur [1.5K]

The answer is 52

x - the number of students in a bus

4 + 9x = 472
9x = 472 - 4
9x = 468
x = 468 / 9
x = 52

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3 years ago

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