What is the phrase to remember the independent variable?

PLEASE HELP ASAP!!! CORRECT ANSWERS ONLY PLEASE!!!

BabaBlast [244]

Hey!

Weight of small box = 5 lb. x no. of small boxes total weight = 5*x

Weight of Large boxes = 12 lb. y no. of large boxes total weight = 12*y

Total weight = 5x + 12y.

Since total weight carrying capacity of shelf is 80, thus the inequality becomes:

5x + 12y Less than or equal to 80.

(a) satisfies the inequality.

(b) satisfies the inequality.

Hope my answer helps!

7 0

3 years ago

In a grid of an auditorium, the vertices of a section of seats are A (40, 20), B (80, 20), C (100, 70),

olganol [36]

Answer:

2500

Step-by-step explanation:

there you go haha

3 0

2 years ago

Sharon pays $98.75 for twenty-five 14-ounce boxes of Yummy flakes cereal. How much does on box of cereal cost

Viefleur [7K]

Answer:

$3.95

Step-by-step explanation:

5 0

3 years ago

Read 2 more answers

Is the dolphin deeper than point C or point E?

Pepsi [2]

Point E is correct answer

7 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