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Vesna [10]
4 years ago
10

Break apart 54 to solve 54 diveded into 6

Mathematics
1 answer:
Elina [12.6K]4 years ago
5 0

Answer:

9

Step-by-step explanation:

First, break apart 54. You can break it into 18+18+18.

Divide each 18 by 6 then add all of your quotients. 3+3+3=9.

Your answer is 9.

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According to a Pew Research Center study, in May 2011, 35% of all American adults had a smart phone (one which the user can use
taurus [48]

Answer:

z=\frac{0.4 -0.35}{\sqrt{\frac{0.35(1-0.35)}{300}}}=1.82  

p_v =P(z>1.82)=0.034  

So the p value obtained was a very low value and using the significance level given \alpha=0.1 we have p_v so we can conclude that we have enough evidence to reject the null hypothesis

And the best conclusion would be:

There is enough evidence to show that more than 35% of community college students own a smart phone (Pvalue = 0.034).

And for the second case the correct system of hypothesis is:

H0: p = 0.078; Ha: p > 0.078

Step-by-step explanation:

Data given and notation

n=300 represent the random sample taken

\hat p=\frac{120}{300}=0.4 estimated proportion of college students that have a smart phone

p_o=0.35 is the value that we want to test

\alpha=0.1 represent the significance level

Confidence=90% or 0.90

z would represent the statistic (variable of interest)

p_v represent the p value (variable of interest)  

Concepts and formulas to use  

We need to conduct a hypothesis in order to test the claim that the proportion is >0.35.:  

Null hypothesis:p\leq 0.35  

Alternative hypothesis:p > 0.35  

When we conduct a proportion test we need to use the z statistic, and the is given by:  

z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}} (1)  

The One-Sample Proportion Test is used to assess whether a population proportion \hat p is significantly different from a hypothesized value p_o.

Calculate the statistic  

Since we have all the info requires we can replace in formula (1) like this:  

z=\frac{0.4 -0.35}{\sqrt{\frac{0.35(1-0.35)}{300}}}=1.82  

Statistical decision  

It's important to refresh the p value method or p value approach . "This method is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.  

The significance level provided \alpha=0.1. The next step would be calculate the p value for this test.  

Since is a right tailed test the p value would be:  

p_v =P(z>1.82)=0.034  

So the p value obtained was a very low value and using the significance level given \alpha=0.1 we have p_v so we can conclude that we have enough evidence to reject the null hypothesis

And the best conclusion would be:

There is enough evidence to show that more than 35% of community college students own a smart phone (Pvalue = 0.034).

And for the second case the correct system of hypothesis is:

H0: p = 0.078; Ha: p > 0.078

4 0
3 years ago
Estimate 196 and 482
geniusboy [140]
When you estimate you round to the nearest whole number (simply to make it easier.) Depending on what place you need to estimate to (like hundreds, or just tens..)

Let's turn 196 into 200 and 482 into 480

200+480 = 680

Hope this hlps!
3 0
4 years ago
Read 2 more answers
I NEED ASAP PLS ONLY RIGHT ANSWERS!!! (24 pts!)
tigry1 [53]
C would be the correct answer with the given information, because with a constant 6% inflation rate the cost of tuition would total in the given years to $99,511

8 0
3 years ago
Read 2 more answers
Find the length of diagonal AV in a three-dimensional figure.​
katovenus [111]

Answer:

C. 13

Step-by-step explanation:

Apply pythagorean theorem.

Thus:

Diagonal AV = √(12² + 5²)

AV = √(144 + 25)

AV = √169

AV = 13 cm

6 0
3 years ago
In a sample of 7500 new iphones, 18 were found to be defective. at this rate, how many defective iphones would be expected in a
viva [34]
In 7500 iPhones, 18 were found defective. 
18 of 7500 is 0.24%
0.24% of iPhones were defective.

Using the information above:
In a sample of 60000 new iPhones, 144 would be found defective.
144 of 60000 is 0.24%

Answer: 144 defective iPhones would be expected in a sample of 60,000.

Hope this helps! Good luck. :-)
8 0
4 years ago
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