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Llana [10]
2 years ago
15

Turn these phrases into algebraic expressions (PLEASE HELP)

Mathematics
2 answers:
Aleks [24]2 years ago
8 0
2x-7
7x+2
x-9
4x+9
4x-9
x/4
Lorico [155]2 years ago
7 0
2x-7
7x+2
X-9
4x+9
9-4x
1/4x

Here is a guide :
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This table shows information about two occupations.
ycow [4]

yes it can be possible to do

5 0
3 years ago
Evaluate the following expression when x = 1/2 and y = -3.<br><br><br> -4xy
solong [7]

Answer:6

Step-by-step explanation:

-4xy

-4(1/2)(-3)

-4(-1.5)

6

3 0
2 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
SOMEONE PLEASE HELP ME i really need this one
jek_recluse [69]

Answer:

Wait are u trying to find the area of a shape???

Step-by-step explanation:

8 0
3 years ago
Which statement describes the expression of 16 + 1/2 * 8 - 6
HACTEHA [7]
16 added to the product of 1/2 times 8, and then subtracted by 6.

16+4-6=20-6=14
4 0
3 years ago
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