Which of the following equations could be used to find the area of the rhombus below?

Which statement holds true for a skewed histogram showing a distribution of the weights of students in a class?

Vanyuwa [196]

Among the choices, the statement which describes a skewed histogram showing a distribution of the weights of students in a class is:

"The nature of the skew can be verified by the position of the mean with respect to the mode."

The histogram is skewed to the right if the mean is less than the mode and the histogram is skewed to the left if the mean is more than the mode.

8 0

3 years ago

Hi, can I get some help with this question? DON'T WORRY ITS JUST A ASSIGNMENT. NOT A TEST. Thank you

Alchen [17]

Answer:

d.

Step-by-step explanation:

i looked at the graph!

3 0

4 years ago

What is 4.06 increased by 17? please help me

Dafna1 [17]

Answer:

21.06

Step-by-step explanation:

4.06 increased by 17

4.06 + 17

21.06

4 0

3 years ago

Write a mathematical sentence that expresses the information given below. Use t as your variable name. If necessary:Susanne chec

Feliz [49]

Answer:

$t+8>82^{\circ}$

Explanation:

We were given that:

Let "temperature" be represented by "t"

Suzan checks the temperature at 11:00. If the temperature rises by 8 more degrees, it will break the record high temperature for the day which is 82 degrees

This information is mathematically represented as:

$t+8>82^{\circ}$

7 0

1 year ago

Location is known to affect the number, of a particular item, sold by an auto parts facility. Two different locations, A and B,

Mama L [17]

We have two samples, A and B, so we need to construct a 2 Samp T Int using this formula:

$\displaystyle \overline {x}_1 - \overline {x}_2 \ \pm \ t^{*} \sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2} }$

In order to use t*, we need to check conditions for using a t-distribution first.

Random for both samples -- NOT STATED in the problem ∴ proceed with caution!
Independence for both samples: 130 < all items sold at Location A; 180 < all items sold at Location B -- we can reasonably assume this is true
Normality: CLT is not met; n < 30 for both locations A and B ∴ proceed with caution!

Since 2/3 conditions aren't met, we can still proceed with the problem but keep in mind that the results will not be as accurate until more data is collected or more information is given in the problem.

Solve for t*:

We need the tail area first.

$\displaystyle \frac{1-.9}{2}= .05$

Next we need the degree of freedom.

The degree of freedom can be found by subtracting the degree of freedom for A and B.

The general formula is df = n - 1.

df for A: 13 - 1 = 12
df for B: 18 - 1 = 17
df for A - B: |12 - 17| = 5

Use a calculator or a t-table to find the corresponding t-score for df = 5 and tail area = .05.

t* = -2.015

Now we can use the formula at the very top to construct a confidence interval for two sample means.

$\overline {x}_A=39$
$s_A=8$
$n_A=13$
$\overline {x}_B = 55$
$s_B=2$
$n_B=18$
$t^{*}=-2.015$

Substitute the variables into the formula: $\displaystyle \overline {x}_1 - \overline {x}_2 \ \pm \ t^{*} \sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2} }$ .

$39-55 \ \pm \ -2.015 \big{(}\sqrt{\frac{(8)^2}{13} +\frac{(2)^2}{18} } } \ \big{)}$

Simplify this expression.

$-16 \ \pm \ -2.015 (\sqrt{5.1453} \ )$
$-16 \ \pm \ 3.73139$

Adding and subtracting 3.73139 to and from -16 gives us a confidence interval of:

$(-20.5707,-11.4293)$

If we want to interpret the confidence interval of (-20.5707, -11.4293), we can say...

We are 90% confident that the interval from -20.5707 to -11.4293 holds the true mean of items sold at locations A and B.

5 0

2 years ago