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3 years ago

12

Janice has 12 sugar cubes, and Jeremy has 16 sugar cubes. Which sentence correctly compares the numbers of sugar cubes Janice an

d Jeremy have?

2 answers:

Volgvan3 years ago

8 0

What are the sentence options?

Lunna [17]3 years ago

7 0

Answer:

what are the sentence options?

Step-by-step explanation:

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Independent random samples of vehicles traveling past a given point on an interstate highway have been observed on monday versus

quester [9]

Hi!

To compare this two sets of data, you need to use a t-student test:

You have the following data:

-Monday n1=16; <span>x̄1=59,4 mph; s1=3,7 mph

-Wednesday n2=20; </span>x̄2=56,3 mph; s2=4,4 mph

You need to calculate the statistical t, and compare it with the value from tables. If the value you obtained is bigger than the tabulated one, there is a statistically significant difference between the two samples.

$t= \frac{X1-X2}{ \sqrt{ \frac{(n1-1)* s1^{2}+(n2-1)* s2^{2} }{n1+n2-2}} * \sqrt{ \frac{1}{n1}+ \frac{1}{n2}} } =2,2510$

To calculate the degrees of freedom you need to use the following equation:

$df= \frac{ (\frac{ s1^{2}}{n1} + \frac{ s2^{2}}{n2})^{2}}{ \frac{(s1^{2}/n1)^{2}}{n1-1}+ \frac{(s2^{2}/n2)^{2}}{n2-1}}=33,89$ ≈34

The tabulated value at 0,05 level (using two-tails, as the distribution is normal) is 2,03. https://www.danielsoper.com/statcalc/calculator.aspx?id=10

So, as the calculated value is higher than the critical tabulated one, we can conclude that the average speed for all vehicles was higher on Monday than on Wednesday.

5 0

3 years ago

Calculate the Average of...

lesantik [10]

You add them and then divide by how many numbers there are.

a. (3+43)/2 = 23
b. 0
c. -1

3 0

3 years ago

Read 2 more answers

The International Air Transport Association surveys business travelers to develop quality ratings for transatlantic gateway airp

motikmotik

Answer:

The 95% confidence interval for the population mean rating is (5.73, 6.95).

Step-by-step explanation:

We start by calculating the mean and standard deviation of the sample:

$M=\dfrac{1}{n}\sum_{i=1}^n\,x_i\\\\\\M=\dfrac{1}{50}(6+4+6+. . .+6)\\\\\\M=\dfrac{317}{50}\\\\\\M=6.34\\\\\\s=\sqrt{\dfrac{1}{n-1}\sum_{i=1}^n\,(x_i-M)^2}\\\\\\s=\sqrt{\dfrac{1}{49}((6-6.34)^2+(4-6.34)^2+(6-6.34)^2+. . . +(6-6.34)^2)}\\\\\\s=\sqrt{\dfrac{229.22}{49}}\\\\\\s=\sqrt{4.68}=2.16\\\\\\$

We have to calculate a 95% confidence interval for the mean.

The population standard deviation is not known, so we have to estimate it from the sample standard deviation and use a t-students distribution to calculate the critical value.

The sample mean is M=6.34.

The sample size is N=50.

When σ is not known, s divided by the square root of N is used as an estimate of σM:

$s_M=\dfrac{s}{\sqrt{N}}=\dfrac{2.16}{\sqrt{50}}=\dfrac{2.16}{7.071}=0.305$

The degrees of freedom for this sample size are:

$df=n-1=50-1=49$

The t-value for a 95% confidence interval and 49 degrees of freedom is t=2.01.

The margin of error (MOE) can be calculated as:

$MOE=t\cdot s_M=2.01 \cdot 0.305=0.61$

Then, the lower and upper bounds of the confidence interval are:

$LL=M-t \cdot s_M = 6.34-0.61=5.73\\\\UL=M+t \cdot s_M = 6.34+0.61=6.95$

The 95% confidence interval for the mean is (5.73, 6.95).

4 0

3 years ago

How many three-letter "words" can be made from 9 letters "fghijklmn" if repetition of letters (a) is allowed?

11Alexandr11 [23.1K]

729. You just do 9 to the power of 3 or 9^3 and you will get 729. If it is not allowed, the answer is 504.(9 x 8 x 7)

8 0

3 years ago

F(x) = -7x2 + 9x - 15 and g(x) = -12x2 – 15x + 7

djyliett [7]

Answer:

${5x}^{2} + 6x = - 8$

Step-by-step explanation:

This question is asking you to insert two equations, f(x) and g(x), into the equation g(x) - f(x). This will lead to the equation below.

${ - 12x}^{2} - 15x + 7 - ({ - 7)x}^{2} + 9x - 15$

Now we can solve for the equation.

${ - 5x}^{2} - 6x - 8 = 0 \\ {5x}^{2} + 6x = - 8$

3 0

2 years ago