Evaluate 4(x+3)(x+1)/(x+5)(x-5) x=3

Elden [556K]

The nearest degree would be 40

4 0

3 years ago

A stone is dropped from the top of a tower. The height h in meters from where the stone is dropped to the ground is given by the

CaHeK987 [17]

Wag umasa sa brainly

Step-by-step explanation:

Labot

5 0

2 years ago

Complete the equation: 96 in = yd.

disa [49]

Answer:

2.666666667

Step-by-step explanation:

12in=1ft

3ft=1yd

96/12=8

8/3=2.666666667

3 0

3 years ago

4.One attorney claims that more than 25% of all the lawyers in Boston advertise for their business. A sample of 200 lawyers in B

AleksAgata [21]

Answer:

$z=\frac{0.315 -0.25}{\sqrt{\frac{0.25(1-0.25)}{200}}}=2.123$

$p_v =P(Z>2.123)=0.0169$

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, and we can said that at 5% of significance the proportion of lawyers had used some form of advertising for their business is significantly higher than 0.25 or 25% .

Step-by-step explanation:

1) Data given and notation

n=200 represent the random sample taken

X=63 represent the lawyers had used some form of advertising for their business

$\hat p=\frac{63}{200}=0.315$ estimated proportion of lawyers had used some form of advertising for their business

$p_o=0.25$ is the value that we want to test

$\alpha=0.05$ represent the significance level

Confidence=95% or 0.95

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 more than 25% of all the lawyers in Boston advertise for their business:

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 $p_o$ .

3) Calculate the statistic

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

$z=\frac{0.315 -0.25}{\sqrt{\frac{0.25(1-0.25)}{200}}}=2.123$

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

Since is a right tailed test the p value would be:

$p_v =P(Z>2.123)=0.0169$

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, and we can said that at 5% of significance the proportion of lawyers had used some form of advertising for their business is significantly higher than 0.25 or 25% .

8 0

3 years ago

Find -7a-6B (Picture provided)

sertanlavr [38]

Answer:

Option c

$-7A-6B=\left(\begin{array}{ccc}40&-106&-32\\-46&43&25\\82&-62&-18\end{array}\right)$

Step-by-step explanation:

First multiply matrix A by -7.

Then multiply the matrix B by -6.

When multiplying a matrix by a number x you must multiply each element of the matrix by x.

Then we perform operation -7A

$-7\left(\begin{array}{ccc}-10&10&8\\-2&-1&5\\-4&8&-6\end{array}\right)=\left(\begin{array}{ccc}70&-70&-56\\14&7&-35\\28&-56&42\end{array}\right)$

Now we perform operation -6B

$-6\left(\begin{array}{ccc}5&6&-4\\10&-6&-10\\-9&1&10\end{array}\right)=\left(\begin{array}{ccc}-30&-36&24\\-60&36&60\\54&-6&-60\end{array}\right)$

Now we add the resulting matrices

$\left(\begin{array}{ccc}70&-70&-56\\14&7&-35\\28&-56&42\end{array}\right)+\left(\begin{array}{ccc}-30&-36&24\\-60&36&60\\54&-6&-60\end{array}\right)=\left(\begin{array}{ccc}40&-106&-32\\-46&43&25\\82&-62&-18\end{array}\right)$

7 0

2 years ago