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Anvisha [2.4K]
3 years ago
13

A concrete slab 4 inches deep will be poured for the floor of greenhouse A. How many cubic feet of concretar are needed for the

floor
Mathematics
1 answer:
Mars2501 [29]3 years ago
8 0

This question is incomplete

Complete Question

Consider greenhouse A with floor dimensions w = 16 feet , l = 18 feet.

A concrete slab 4 inches deep will be poured for the floor of greenhouse A. How many cubic feet of concrete are needed for the floor?

Answer:

96 cubic feet

Step-by-step explanation:

The volume of the floor of the green house = Length × Width × Height

We convert the dimensions in feet to inches

1 foot = 12 inches

For width

1 foot = 12 inches

16 feet = x

Cross Multiply

x = 16 × 12 inches

x = 192 inches

For length

1 foot = 12 inches

18 feet = x

Cross Multiply

x = 18 × 12 inches

x = 216 inches

The height or depth = 4 inches deep

Hence,

Volume = 192 inches × 216 inches × 4 inches

= 165888 cubic inches

From cubic inches to cubic feet

1 cubic inches = 0.000578704 cubic foot

165888 cubic inches = x

Cross Multiply

x = 16588 × 0.000578704 cubic foot

x = 96 cubic feet

Therefore, 96 cubic feet of concrete is needed for the floor

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0.001665973

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Step-by-step explanation:

If you borrow $120,000 at an APR of 7% for 25 years, you will pay $848.13 per month. If you

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Vx-w=2 <br> rearrange it and make x the subject
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3 years ago
A genetic experiment involving peas yielded one sample of offspring consisting of 420 green peas and 174 yellow peas. Use a 0.01
slavikrds [6]

Answer:

a) z=\frac{0.293 -0.23}{\sqrt{\frac{0.23(1-0.23)}{594}}}=3.649  

b) For this case we need to find a critical value that accumulates \alpha/2 of the area on each tail, we know that \alpha=0.01, so then \alpha/2 =0.005, using the normal standard table or excel we see that:

z_{crit}= \pm 2.58

Since the calculated value is higher than the critical value we have enough evidence to reject the null hypothesis at 1% of significance.

Step-by-step explanation:

Data given and notation

n=420+174=594 represent the random sample taken

X=174 represent the number of yellow peas

\hat p=\frac{174}{594}=0.293 estimated proportion of yellow peas

p_o=0.23 is the value that we want to test

\alpha=0.01 represent the significance level

Confidence=99% or 0.99

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 true proportion of yellow peas is 0.23:  

Null hypothesis:p=0.23  

Alternative hypothesis:p \neq 0.23  

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{0.293 -0.23}{\sqrt{\frac{0.23(1-0.23)}{594}}}=3.649  

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=0.05. 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>3.649)=0.00026  

So the p value obtained was a very low value and using the significance level given \alpha=0.05 we have p_v so we can conclude that we have enough evidence to reject the null hypothesis.

b) Critical value

For this case we need to find a critical value that accumulates \alpha/2 of the area on each tail, we know that \alpha=0.01, so then \alpha/2 =0.005, using the normal standard table or excel we see that:

z_{crit}= \pm 2.58

Since the calculated value is higher than the critical value we have enough evidence to reject the null hypothesis at 1% of significance.

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3 years ago
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Answer:

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