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mr Goodwill [35]

2 years ago

11

The base BC of an equilateral triangle ABC lies on y-axis. The coordinates of points C are (0, -3). The origin is the mid –point

of the base. Find the coordinates of the points A and B. Also find the coordinates of another point D such that BACD is a rhombus.

1 answer:

Kobotan [32]2 years ago

6 0

Answer

Point b is (0,0) Point a is (3,-2) Im not doing the second part

Step-by-step explanation:

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Answer:

<h2>2. Because 25<12, this equation can not be true.</h2>

Step-by-step explanation:

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3 years ago

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Answer:

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Step-by-step explanation:

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3 years ago

Radio stations use electromagnetic waves for broadcasting. The chart shows different frequencies of waves used by radio stations

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2 years ago

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In the past decades there have been intensive antismoking campaigns sponsored by both federal and private agencies. In one study

DIA [1.3K]

Answer:

$z=\frac{0.384-0.362}{\sqrt{0.374(1-0.374)(\frac{1}{4276}+\frac{1}{3908})}}=2.055$

The p value can be calculated from the alternative hypothesis with this probability:

$p_v =2*P(Z>2.055)=0.0399$

And the best option for this case would be:

C. between 0.01 and 0.05.

Step-by-step explanation:

Information provided

$X_{1}=1642$ represent the number of smokers from the sample in 1995

$X_{2}=1415$ represent the number of smokers from the sample in 2010

$n_{1}=4276$ sample from 1995

$n_{2}=3908$ sample from 2010

$p_{1}=\frac{1642}{4276}=0.384$ represent the proportion of smokers from the sample in 1995

$p_{2}=\frac{1415}{3908}=0.362$ represent the proportion of smokers from the sample in 2010

$\hat p$ represent the pooled estimate of p

z would represent the statistic

$p_v$ represent the value for the pvalue

System of hypothesis

We want to test the equality of the proportion of smokers and the system of hypothesis are:

Null hypothesis: $p_{1} = p_{2}$

Alternative hypothesis: $p_{1} \neq p_{2}$

The statistic is given by:

$z=\frac{p_{1}-p_{2}}{\sqrt{\hat p (1-\hat p)(\frac{1}{n_{1}}+\frac{1}{n_{2}})}}$ (1)

Where $\hat p=\frac{X_{1}+X_{2}}{n_{1}+n_{2}}=\frac{1642+1415}{4276+3908}=0.374$

Replacing the info given we got:

$z=\frac{0.384-0.362}{\sqrt{0.374(1-0.374)(\frac{1}{4276}+\frac{1}{3908})}}=2.055$

The p value can be calculated from the alternative hypothesis with this probability:

$p_v =2*P(Z>2.055)=0.0399$

And the best option for this case would be:

C. between 0.01 and 0.05.

7 0

2 years ago