2 y -94 24 74 18 -54 12 What is the x-intercept of the line?

Find the product of -5 1/3 • 4 2/5

fomenos

Answer:

Simplify the expression.

Exact Form: -352/15

Decimal Form: -23.46

Mixed Number Form: -23 7/15

Step-by-step explanation:

7 0

3 years ago

What is 49 times 49! Plz help me.

Natasha_Volkova [10]

2401 I had that question a few years ago

7 0

4 years ago

The National Transportation Safety Board publishes statistics on the number of automobile crashes that people in various age gro

ra1l [238]

Answer:

a) Null hypothesis: $p\leq 0.12$

Alternative hypothesis: $p > 0.12$

b) $z_{\alpha}=1.64$

c) $z=\frac{0.134 -0.12}{\sqrt{\frac{0.12(1-0.12)}{1000}}}=1.362$

d) $p_v =P(z>1.362)=0.0866$

e) For this case since the statistic is lower than the critical value and the p value higher than the significance level we have enough evidence to FAIL to reject the null hypothesis so then we don't have information to conclude that the true proportion is higher than 0.12

Step-by-step explanation:

Information given

n=1000 represent the random sample selected

X=134 represent the number of young drivers ages 18 – 24 that had an accident

$\hat p=\frac{134}{1000}=0.134$ estimated proportion of young drivers ages 18 – 24 that had an accident

$p_o=0.12$ is the value that we want to verify

$\alpha=0.05$ represent the significance level

Confidence=95% or 0.95

z would represent the statistic

$p_v{/tex} represent the p valuePart aWe want to verify if the population proportion of young drivers, ages 18 – 24, having accidents is greater than 12%: Null hypothesis:[tex]p\leq 0.12$

Alternative hypothesis: $p > 0.12$

The statistic would be given by:

$z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}$ (1)

Part b

For this case since we are conducting a right tailed test we need to find a critical value in the normal standard distribution who accumulates 0.05 of the area in the right and we got:

$z_{\alpha}=1.64$

Part c

For this case the statistic would be given by:

$z=\frac{0.134 -0.12}{\sqrt{\frac{0.12(1-0.12)}{1000}}}=1.362$

Part d

The p value can be calculated with the following probability:

$p_v =P(z>1.362)=0.0866$

Part e

For this case since the statistic is lower than the critical value and the p value higher than the significance level we have enough evidence to FAIL to reject the null hypothesis so then we don't have information to conclude that the true proportion is higher than 0.12

8 0

3 years ago

Please follow the directions. Don't round immediate calculations. Please try to get it right

marin [14]

Answer:

well without rounding anything, I got 511.9999999992 people

Step-by-step explanation:

so the ratio I used was 36 people for every 225 people( 225 because in total he asked 225 people) then I divided 3,200 into 225 to see how many times It would fit, the answer was 14.2222222222 then I multiplied that by 36 and my answer was 511.9999999992 people. I hope this helps, I tried to be as accurate as possible.

4 0

3 years ago

The box plots below show student grades on the most recent exam compared to overall grades in the class: two box plots shown. Th

finlep [7]

Question:

The box plots below show student grades on the most recent exam compared to overall grades in the class: two box plots shown. The top one is labeled Class. Minimum at 70, Q1 at 74, median at 83, Q3 at 92, maximum at 100. The bottom box plot is labeled Exam. Minimum at 60, Q1 at 81, median at 87, Q3 at 91, maximum at 95. Which of the following best describes the information about the medians? (1 point) Group of answer choices

•The exam outlier at 60 makes the IQR narrower and the median higher.

•The class data is more evenly spread, which pulls its median down.

•The class median is lower than the exam median.

•The class Q3 is higher than the exam Q3.

Answer:

•The class Q3 is higher than the exam Q3.

Step-by-step explanation:

In the question we have two box plots

a) Box plots for class is given as

Minimum at 70, Q1 at 74, median at 83, Q3 at 92,

b) Box plots for exam is given as Minimum at 60, Q1 at 81, median at 87, Q3 at 91, maximum at 95.

When we compare the median of the class which is 83 and the median of the exam plots 87, we can see that the median of the exam is higher than the median of the class, hence the option

"•The class median is lower than the exam median". is wrong.

From the options given, the only correct option which best describes the information about the medians is

"•The class Q3 is higher than the exam Q3".

This is because when we compare the Q3 of the class which is 92 to the Q3 of the box which is 91 , we can see that the class Q3 is higher than the box Q3 hence the option is correct.

8 0

3 years ago