Solving systems of equations using substitution with both equations in y=mx+b

jeyben [28]

$\bf \begin{array}{llll}
b^a=64\qquad \boxed{64=4^3}\qquad b^a=4^3
\\\\\\
a^b=81\qquad \boxed{81=3^4}\qquad a^b=3^4
\end{array} \qquad 
\begin{cases}
a=3\\
b=4
\end{cases}
\\\\\\
a+b\implies 3+4$

3 0

2 years ago

HURRRY PLEASE. Find the focus and directrix of the parabola with the equation y2=18x

Lady_Fox [76]

Been a while but I'm gonna try..! Okay!
For the Focus use the vertex to find it. (9/2 , 0)
For the directrix I believe it is. x = -9/2

7 0

2 years ago

6 more than 3 times what number squares is 81

Lisa [10]

When you write that equation, you get 3x^2+6=81.

Solving for x:

Subtract by 6 on both sides of the equation to get 3x^2=75.

Then, divide by 3 on both sides of the equation to get x^2=25.

Finally, do the square root on both sides of the equation to get x=5.

The number is 5.

6 0

3 years ago

Darnel has pulled 4 brown marbles and 10 other marbles from a large bag. Based on experimental probability, how many of the next

saw5 [17]

Answer:

8 brown marbles

Step-by-step explanation:

Daniel has pulled 4 brown out of 14.

The number of marbles doubled, so the number of brown is expected to double:

4*(28/14) = 4*2 = 8

7 0

2 years ago

In a survey of 1000 women age 22 to 35 who work full time, 540 indicated that they would be willing to give up some personal tim

hram777 [196]

Answer:

$z=\frac{0.540-0.5}{\sqrt{\frac{0.5(1-0.5)}{1000}}}=2.53$

$p_v =P(Z>2.53)=0.0057$

So the p value obtained was a very low value and using the significance level given $\alpha=0.01$ we have $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 1% of significance the true proportion is not significantly higher than 0.5.

Step-by-step explanation:

1) Data given and notation

n=1000 represent the random sample taken

X=540 represent the people indicated that they would be willing to give up some personal time in order to make more money

$\hat p=\frac{540}{1000}=0.54$ estimated proportion of people indicated that they would be willing to give up some personal time in order to make more money

$p_o=0.5$ 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 majority of women age 22 to 35 who work full-time would be willing to give up some personal time for more money :

Null hypothesis: $p\leq 0.5$

Alternative hypothesis: $p > 0.5$

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 $p_o$ .

<em>Check for the assumptions that he sample must satisfy in order to apply the test </em>

a)The random sample needs to be representative: On this case the problem no mention about it but we can assume it.

b) The sample needs to be large enough

$np_o =100*0.5=500>10$

$n(1-p_o)=1000*(1-0.5)=500>10$

3) Calculate the statistic

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

$z=\frac{0.540-0.5}{\sqrt{\frac{0.5(1-0.5)}{1000}}}=2.53$

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 right tailed test the p value would be:

$p_v =P(Z>2.53)=0.0057$

So the p value obtained was a very low value and using the significance level given $\alpha=0.01$ we have $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 1% of significance the true proportion is not significantly higher than 0.5.

7 0

2 years ago