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Brums [2.3K]
3 years ago
12

Emmanuel bought 28 lb of potting soil this week. This amount is 4 lb. more than twice the amount of potting soil he bought last

week. Answer the following questions to find the number of pounds of soil Emmanuel purchased last week.
(b) Let p represent the unknown information. Write an equation to model the problem.
show your work
Mathematics
2 answers:
musickatia [10]3 years ago
7 0

Answer:

7 (28lb ÷ 4lb = 7lb)

Step-by-step explanation:

28÷4= p

makvit [3.9K]3 years ago
3 0
An equation for this:

28 * 2 + 4 = p

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5.28 A manufacturer knows that on average 20% of the electric toasters produced require repairs within 1 year after they are sol
ahrayia [7]

Answer:

A) probability that at least x of them will require repairs is less than 0.5 is 4

Step-by-step explanation:

sample size, n = 20

p = success probability of toaster = 0.2

q = failure probability of toaster =1- p = 1- 0.2 =0.8

mean, , μ = n * p = 20 * 0.2 = 4

standard deviation, \sigma = \sqrt{npq} = \sqrt{(20 * 0.2 *0.8)} = 1.79

a) the probability that at least some of them require repairs < 0.5.

for p \geq 0.5 value of z = 0

we know z  =\frac{x -\mu}{\sigma}

0 = \frac{X -4}{1.79} from z table

X =  4

B)  SECOND PART IS INCOMPLETE { DATA IS INCOMPLETE}

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Answer:

I need this same question

Step-by-step explanation:

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Naya [18.7K]

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Answer:

  135.1 cm

Step-by-step explanation:

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The length of one of the braces is about 135.1 cm.

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Suppose we are testing people to see if the rate of use of seat belts has changed from a previous value of 88%. Suppose that in
Andreas93 [3]

Answer:

a) We would expect to see 500*0.88=440

b) z=\frac{0.9 -0.88}{\sqrt{\frac{0.88(1-0.88)}{500}}}=1.376  

p_v =2*P(Z>1.376)=0.167  

So the p value obtained was a very high value and using the significance level assumed \alpha=0.05 we have p_v>\alpha so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 5% of significance the true proportion is not significant different from 0.9.

The p value is a criterion to decide if we reject or not the null hypothesis, when p_v we reject the null hypothesis in other case we FAIL to reject the null hypothesis. And represent the "probability of obtaining the observed results of a test, assuming that the null hypothesis is correct".  

Step-by-step explanation:

Data given and notation

n=500 represent the random sample taken

X=450 represent the people that have the seat belt fastened

\hat p=\frac{450}{500}=0.9 estimated proportion of people that have the seat belt fastened

p_o=0.88 is the value that we want to test

\alpha represent the significance level

z would represent the statistic (variable of interest)

p_v{/tex} represent the p value (variable of interest)  Part aWe would expect to see 500*0.88=440Part bConcepts and formulas to use  We need to conduct a hypothesis in order to test the claim that the true proportion changes fro m 0.88.:  Null hypothesis:[tex]p=0.88  

Alternative hypothesis:p \neq 0.88  

When we conduct a proportion test we need to use the z statisitc, 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{0.9 -0.88}{\sqrt{\frac{0.88(1-0.88)}{500}}}=1.376  

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 assumed is \alpha=0.05. The next step would be calculate the p value for this test.  

Since is a bilateral test the p value would be:  

p_v =2*P(Z>1.376)=0.167  

So the p value obtained was a very high value and using the significance level assumed \alpha=0.05 we have p_v>\alpha so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 5% of significance the true proportion is not significant different from 0.9.

The p value is a criterion to decide if we reject or not the null hypothesis, when p_v we reject the null hypothesis in other case we FAIL to reject the null hypothesis. And represent the "probability of obtaining the observed results of a test, assuming that the null hypothesis is correct".  

8 0
3 years ago
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