Ask question

Ask question

All categories

3 years ago

5

A jar contains 3 pennies, 5 nickels, 4 dimes, and 6 quarters. If a coin is selected at random, what is the probability of select

ing a penny?

2 answers:

bagirrra123 [75]3 years ago

7 0

The probability of selecting a penny would be 1/6

klio [65]3 years ago

5 0

Answer:

Step-by-step explanation:

3 out of 18 chance

You might be interested in

HURRYYYYY<br> What is the value of the expression 4’4?

WINSTONCH [101]

Answer:

256

Step-by-step explanation:

Raise 4 to the power of 4

Basically, multiply 4 by itself four times. :)

4 0

3 years ago

Rachel's chocolate bar is 43% cocoa. If the weight of the chocolate bar is 68 grams, how many grams of cocoa does it contain? Ro

Yanka [14]

Answer:

29.24 grams

Step-by-step explanation:

It is given that,

Rachel's chocolate bar is 43% cocoa

The weight of the chocolate bar is 68 grams

We need to find how many grams of cocoa does it contain. It means it will contain 43% of 68 grams. It can be solved as follows :

$C=\dfrac{43}{100}\times 68\\\\C=29.24\ \text{grams}$

Hence, it will have 29.24 grams of cocoa.

7 0

3 years ago

Please Can Someone Tell Me The Area??

torisob [31]

Answer:

66 I am pretty sure

Step-by-step explanation:

6 x 2 divided by 2 = 6

6 + 6 = 12 x 10 divided by 2 = 60

add them together

60 + 6 = 66

3 0

2 years ago

The log: 20 can be written as

adoni [48]

Answer:

10+10

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers

Nitrates are groundwater contaminants derived from fertilizer, septic tank seepage and other sewage. Nitrate poisoning is partic

o-na [289]

Answer:

a) $n=\frac{0.5(1-0.5)}{(\frac{0.02}{1.96})^2}=2401$

And rounded up we have that n=2401

b) $n=\frac{0.07(1-0.07)}{(\frac{0.02}{1.96})^2}=625.22$

And rounded up we have that n=626

And we see that if we have previous info about p the sample size varies a lot.

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".

Solution to the problem

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:

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

Part a

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 $ME =\pm 0.02$ 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)

Since we don't have prior information about the population proportion we can use ase estimator $\hat p =0.5$ . And replacing into equation (b) the values from part a we got:

$n=\frac{0.5(1-0.5)}{(\frac{0.02}{1.96})^2}=2401$

And rounded up we have that n=2401

Part b

For thi case the estimator for p is $\hat p =0.07$

$n=\frac{0.07(1-0.07)}{(\frac{0.02}{1.96})^2}=625.22$

And rounded up we have that n=626

And we see that if we have previous info about p the sample size varies a lot.

4 0

3 years ago