Henlo! <br> how is everyone doing?<br> OK 15 points<br> name all the colors in the rainbow!

If I have 24 jolly ranchers and 64 starbursts what is the largest amount of bags I can fill? How many starbursts will be I’m eac

Rus_ich [418]

Answer:

Largest amount of bags that can be filled: 11

Starbursts in each bag: 3

Step-by-step explanation:

Find the greastest common factor between these two (which is 8) and divide the sum of the jolly ranchers and starbursts-which will be 88-by the greatest common factor. This will leave you with 11 bags that can be filled with 8 candies. To find the starbursts, divide 24 by 8 (3.)

5 0

3 years ago

A motel bills its customers by charging a flat fee for the first night and then adding on a fixed amount for every night thereaf

melamori03 [73]

65 is the flat fee for the first night followed by a charge of 45 a night.

8 0

3 years ago

An online news source reports that the proportion of smartphone owners who use a certain operating system is 0.216. Two simulati

kaheart [24]

Answer:

The center/ mean will almost be equal, and the variability of simulation B will be higher than the variability of simulation A.

Step-by-step explanation:

Solution

Normally, a distribution sample is mostly affected by sample size.

As a rule, sampling error decreases by half by increasing the sample size four times.

In this case, B sample is 2 times higher the A sample size.

Now, the Mean sampling error is affected and is not higher for A.

But it's sample is huge for this, Thus, they are almost equal

Variability of simulation decreases with increase in number of trials. A has less variability.

With increase number of trials, variability of simulation decreases, so A has less variability.

4 0

3 years ago

On any day, the probability of rain is 0.3. The occurrence of rain on any day is independent of the occurrence of rain on any ot

poizon [28]

<h3>Answer: 0.47178 Step-by-step explanation: Find the probability for each p(X=x) up to 5 using the equation: (x-1)C(r-1)*p^r * q^x-r, where x is number of days, p = .3 (prob of rain). q=.7 (prob of not rain), and r=2 (second day of rain). also C means choose. So p(X=1) = 0 p(X=2) = 1C1 * .3^2 * .7^0 = .09 P(X=3) = 2C1 * .3^2 * .7^1 = .126 P(X=4) = 3C1 * .3^2 * .7^2 = .1323 P(X=5) = 4C1 * .3^2 * .7^3 = .12348 Then add all of them up 0+.09+.126+.1323+.12348 = .47178</h3>

3 0

3 years ago

Granola Crunch cereal is packaged in 1 pound boxes. Susan Torres, a quality control analyst who works for the manufacturer of th

sineoko [7]

Answer:

$z=\frac{1.02-1}{\frac{0.04}{\sqrt{40}}}=3.162$

$p_v =2*P(z>3.162)=0.0016$

If we compare the p value and the significance level given $\alpha=0.05$ we see that $p_v$ so we can conclude that we have enough evidence to to reject the null hypothesis, so we can conclude that the true mean is different from 1 pound at 5% of signficance.

Step-by-step explanation:

Data given and notation

$\bar X=1.02$ represent the sample mean

$\sigma=0.04$ represent the population standard deviation

$n=40$ sample size

$\mu_o =1$ represent the value that we want to test

$\alpha=0.05$ represent the significance level for the hypothesis test.

z would represent the statistic (variable of interest)

$p_v$ represent the p value for the test (variable of interest)

State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the true mean is equal to 1 pound or no :

Null hypothesis: $\mu =1$

Alternative hypothesis: $\mu \neq 1$

Since we know the population deviation, is better apply a z test to compare the actual mean to the reference value, and the statistic is given by:

$z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}}$ (1)

z-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".

Calculate the statistic

We can replace in formula (1) the info given like this:

$z=\frac{1.02-1}{\frac{0.04}{\sqrt{40}}}=3.162$

P-value

Since is a two-sided test the p value would be:

$p_v =2*P(z>3.162)=0.0016$

Conclusion

If we compare the p value and the significance level given $\alpha=0.05$ we see that $p_v$ so we can conclude that we have enough evidence to to reject the null hypothesis, so we can conclude that the true mean is different from 1 pound at 5% of signficance.

5 0

3 years ago

Henlo! how is everyone doing? OK 15 points name all the colors in the rainbow!