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Blizzard [7]
3 years ago
12

A child building a tower with Blocks uses 15 for the bottom row. Each row has 2 fewer blocks than the previous row. What formula

would you use to find the number of blocks in the 5th row?
A)blocks in 5th row =15+(15-1)-2
B)blocks in 5th row=15+(5-1)(-2)
C)blocks in 5th row=15+(5+1)(-2)
D)blocks in 5th row=5+(15-1)(-2)
Mathematics
1 answer:
xenn [34]3 years ago
4 0

Answer:

B)blocks in 5th row=15+(5-1)(-2)

Step-by-step explanation:

We are given that

Number of block in bottom row=15

Each row has 2 fewer blocks than the previous row

We have to find the formula would you use to find the number of blocks in the 5th row.

Number of blocks in  second row=15-2=15+1(-2)

Number of blocks in third row=15-2-2=15+2(-2)

Number of blocks in 4th row=15-2-2-2=15+3(-2)

Number of blocks in 5th row=15-2-2-2-2=15+4(-2)

Therefore, number of block in 5th row

15+(5-1)(-2)

This is required formula to find the number of blocks in the 5th row

Hence, option B is true.

B)blocks in 5th row=15+(5-1)(-2)

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Y= .5x.2^x

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A random sample of 150recent donations at a certain
gtnhenbr [62]

Answer:

Null hypothesis:p\geq 0.4  

Alternative hypothesis:p < 0.4  

z=\frac{0.3 -0.4}{\sqrt{\frac{0.4(1-0.4)}{150}}}=-2.5  

p_v =2*P(Z  

If we compare the p value obtained and the significance level given \alpha=0.01 we see that p_v>\alpha so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that the true proportion is not significantly lower than 0.4 or 40% at 1% of significance.  

Step-by-step explanation:

1) Data given and notation  

n=150 represent the random sample taken

X=45 represent the people with type A blood

\hat p=\frac{45}{150}=0.3 estimated proportion of people with type A blood

p_o=0.4 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)  

2) Concepts and formulas to use  

We need to conduct a hypothesis in order to test the claim that the true proportion of people type A blood is less than 0.4:  

Null hypothesis:p\geq 0.4  

Alternative hypothesis:p < 0.4  

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.

3) Calculate the statistic  

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

z=\frac{0.3 -0.4}{\sqrt{\frac{0.4(1-0.4)}{150}}}=-2.5  

4) 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=0.01. 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  

If we compare the p value obtained and the significance level given \alpha=0.01 we see that p_v>\alpha so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that the true proportion is not significantly lower than 0.4 or 40% at 1% of significance.  

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