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natta225 [31]
2 years ago
12

1. Bob has 5 pounds of nails to build 3 identical doghouses. How many pounds of nails will he use for each?​

Mathematics
2 answers:
siniylev [52]2 years ago
7 0

Answer:

0.6 pounds of nails

Step-by-step explanation:

3/5=0.60

valkas [14]2 years ago
3 0

Answer:

Just divide the 5 and 3 to get 1.6, meaning he will use 1.6 pounds per house.

Step-by-step explanation:

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Determine the coordinates of the point $P$ on the line $y=-x+6$ such that $P$ is equidistant from the points $A(10,-10)$ and $O(
lidiya [134]

Answer:

P=(3,3)

Step-by-step explanation:

We need to find a point  P in the line y=-x+6 such that PA=P0 where A=(10,-10)

Observe that the coordinates of P are (x,y)=(x,-x+6) because P is in the line.

Then,

PA=P0\\(x,-x+6)(10,-10)=(x,-x+6)(0,0)\\10x+(-x+6)(-10)=0\\10x+10x-60=0\\20x=60\\x=3

We reeplace  the value of x in the coordinates of P and obtain the point

(3,-3+6)=(3,3)=P

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2 years ago
A buyer went to the market to buy strawberries. He purchased 120 randomly selected strawberries from a vendor who claimed that n
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Answer:

z=\frac{0.333 -0.25}{\sqrt{\frac{0.25(1-0.25)}{120}}}=2.10  

p_v =P(z>2.10)=0.018  

At 5% of significance we can conclude that the true proportion of strawberries damage is higher than 0.25

Step-by-step explanation:

Data given and notation

n=120 represent the random sample taken

X=40 represent the number of strawberries damaged

\hat p=\frac{40}{120}=0.333 estimated proportion of strawberries damaged

p_o=0.25 is the value that we want to test

\alpha represent the significance level

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 no more than 25% of his total harvest of strawberries was damaged.:  

Null hypothesis:p\leq 0.25  

Alternative hypothesis:p > 0.25  

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{0.333 -0.25}{\sqrt{\frac{0.25(1-0.25)}{120}}}=2.10  

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 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>2.10)=0.018  

At 5% of significance we can conclude that the true proportion of strawberries damage is higher than 0.25

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