H + 2.34 = 9.48<br> solve for H

What is -1 11/12 in it’s simplest form

Dafna1 [17]

Answer: - 23/12

decimal -1.916

Step-by-step explanation:

3 0

3 years ago

A man Buy television set at a price exclusive of sales tax for $2124 if sales tax of 12% is changed how much money does the man

Fed [463]

Answer:

2378.88

Step-by-step explanation:

12/100 x 2124

=254.88

2124+254.88=answer

6 0

2 years ago

Which is a solution for the equation 3x + 13 = 73?

sweet [91]

Answer:

x=20

Step-by-step explanation:

73-13=60

60÷3=20

7 0

2 years ago

Read 2 more answers

What are the zeros of the polynomial function f(x) = x^3-10x^2+24x

Ludmilka [50]

Replace F(x) with y
y=x^3-10x^2+24x
To find the roots replace y with 0.
0=x^3-10x^2-24x
Rewrite the equation as
x^3-10x^2+24x=0
Factor the left side
x(x-6)(x-4)=0
The solution is a result of
x=0, x-6=0, and x-4=0

The answer is:
x=0, 6, 4

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

H + 2.34 = 9.48 solve for H