Graph y= 2sin (x/3

Marley loves to draw and color. She bought 6 boxes of crayons. There are 8 crayons in each box. What is the total number of cray

Tanzania [10]

Answer:

48 crayons

Step-by-step explanation:

There are 6 boxes total, each with 8 crayons in them. Notice that if you would count all the crayons, you would add 8 to the total number of crayons each time you counted a box. There are 6 boxes, so that would be equivalent to adding 8 a total of six times. You would use multiplication: 6 x 8 which is 48.

7 0

3 years ago

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The probability of selecting a particular color almond m&m (according to the website) from a bag of m&ms is listed below

earnstyle [38]

Omg was a weird dream last week and then they didn’t have to go back home so we had some fun of them all over tomorrow and they probably won’t get them tomorrow night but so

4 0

3 years ago

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In San Jose a sample of 73 mail carriers showed that 30 had been bitten by an animal during one week. In San Francisco in a samp

dsp73

Answer:

$(0.411-0.7) - 1.96 \sqrt{\frac{0.411(1-0.411)}{73} +\frac{0.7(1-0.7)}{80}}=-0.4401$

$(0.411-0.7) + 1.96 \sqrt{\frac{0.411(1-0.411)}{73} +\frac{0.7(1-0.7)}{80}}=-0.1380$

We are confident at 95% that the difference between the two proportions is between $-0.4401 \leq p_B -p_A \leq -0.1380$

1. -.4401 ≤ p1 - p2 ≤ -.1380

4. The rate of mail carriers being bitten in San Jose is statistically less than the rate San Francisco at α = 5%

Step-by-step explanation:

In San Jose a sample of 73 mail carriers showed that 30 had been bitten by an animal during one week. In San Francisco in a sample of 80 mail carriers, 56 had received animal bites. Is there a significant difference in the proportions? Use a 0.05. Find the 95% confidence interval for the difference of the two proportions. Sellect all correct statements below based on the data given in this problem.

1. -.4401 ≤ p1 - p2 ≤ -.1380

2. -.4401 ≤ p1 - p2 ≤ .1380

3. The rate of mail carriers being bitten in San Jose is statistically greater than the rate San Francisco at α = 5%

4. The rate of mail carriers being bitten in San Jose is statistically less than the rate San Francisco at α = 5%

5. The rate of mail carriers being bitten in San Jose and San Francisco are statistically equal at α = 5%

Solution to the problem

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".

$p_1$ represent the real population proportion for San Jose

$\hat p_1 =\frac{30}{73}=0.411$ represent the estimated proportion for San Jos

$n_1=73$ is the sample size required for San Jose

$p_2$ represent the real population proportion for San Francisco

$\hat p_2 =\frac{56}{80}=0.7$ represent the estimated proportion for San Francisco

$n_2=80$ is the sample size required for San Francisco

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

The population proportion have the following distribution

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

The confidence interval for the difference of two proportions would be given by this formula

$(\hat p_1 -\hat p_1) \pm z_{\alpha/2} \sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1} +\frac{\hat p_2 (1-\hat p_2)}{n_2}}$

For the 95% confidence interval the value of $\alpha=1-0.95=0.05$ and $\alpha/2=0.025$ , with that value we can find the quantile required for the interval in the normal standard distribution.

$z_{\alpha/2}=1.96$

And replacing into the confidence interval formula we got:

$(0.411-0.7) - 1.96 \sqrt{\frac{0.411(1-0.411)}{73} +\frac{0.7(1-0.7)}{80}}=-0.4401$

$(0.411-0.7) + 1.96 \sqrt{\frac{0.411(1-0.411)}{73} +\frac{0.7(1-0.7)}{80}}=-0.1380$

We are confident at 95% that the difference between the two proportions is between $-0.4401 \leq p_B -p_A \leq -0.1380$

Since the confidence interval contains all negative values we can conclude that the proportion for San Jose is significantly lower than the proportion for San Francisco at 5% level.

Based on this the correct options are:

1. -.4401 ≤ p1 - p2 ≤ -.1380

4. The rate of mail carriers being bitten in San Jose is statistically less than the rate San Francisco at α = 5%

8 0

3 years ago

Identify the statement(s) that are both biconditional and false. Three points are not in the same plane if and only if exactly o

Marysya12 [62]

Answer:

Step-by-step explanation:

1) Three points are not in the same plane if and only if exactly one line passes through them.

It is biconditional and false because more than one plane can pass through three points

2)Exactly one line passes through three points if the points are in an infinite number of planes.

It is biconditional and true because three points on same line will be in more than one plane and if they are not in the same plane then connect them and it will form a triangle which is in only one plane

3). an infinite number of planes pass through the same three points if and only if the points are not on the same line.

It is biconditional and false because more than one plane passes through them So, infinite no. of planes passes through them

4). three points on the same line if and only if exactly one plane passes through them.

It is biconditional and false because more than one plane passes through them So, infinite no. of planes passes through them

7 0

3 years ago

Please help QwQ<br> I'll do anything-

evablogger [386]

Answer:

in scientific notation....14×10^12miles

5 0

3 years ago

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Graph y= 2sin (x/3 - π/6)