How to write a decimal as a fraction?

The ages of the people who came to a birthday party are shown below. Which statements must be true according to the histogram? C

spayn [35]

From the histogram it can be seen that
1.) the data is not qualitative. (Qualitative data is typically descriptive data and is not measured by a number but age is measured by a number).
2.) The data is not skewed right. (For a right skewed distribution, the tail of the distribution on the right hand (positive) side is longer than on the left hand side).
3.) There were 31 people at the party. (The total number of people at the party from the histogram is given by the sum of the heights of the bars of the histogram = 5 + 13 + 8 + 3 + 2 = 31).
4.) There were children at the party. (From the histogram, there were 5 poeple aged 20 and below, which we can describe as children).
5.) The most common age of the attendees was not 13 years old. (The mode (i.e. most occuring frequency) of a distribution is represented by the height of the tallest bar of the histogram. From the histogram, the tallest bar represents the ages between 21 to 40)
6.) The majority of the people at the party were less than 61 years old. (From the histogram, there were 5 + 13 + 8 = 26 people below the age of 61 in the party and 3 + 2 = 5 people above the age of 60 in the party).

Therefore, the statements that must be true according to the histogram are
There were 31 people at the party.
There were children at the party.
and
The majority of the people at the party were less than 61 years old.

5 0

3 years ago

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The following data lists the ages of a random selection of actresses when they won an award in the category of Best Actress, al

Valentin [98]

Answer:

a) $p_v =P(t_{(9)}$

The p value is higher than the significance level given 0.01, so then we can conclude that we FAIL to reject the null hypothesis. And we can say that the true difference for Best Actresses is not significantly lower than the mean for Best Actors at 1% of significance.

b) The 99% confidence interval would be given by (-21.469;2.069)

c) We got the same conclusion as part a, sicne the confidence interval contains the value 0, we FAIL to reject the null hypothesis that the difference between the two

Step-by-step explanation:

Part a

Let put some notation

x=actor's age , y = actress's age

x: 58 41 36 36 34 33 48 37 37 43

y: 26 27 34 26 35 29 23 42 30 34

The system of hypothesis for this case are:

Null hypothesis: $\mu_y- \mu_x \geq 0$

Alternative hypothesis: $\mu_y -\mu_x$

The first step is calculate the difference $d_i=y_i-x_i$ and we obtain this:

d: -32, -14, -2, -10, 1, -4, -25, 5, -7, -9

The second step is calculate the mean difference

$\bar d= \frac{\sum_{i=1}^n d_i}{n}= -9.7$

The third step would be calculate the standard deviation for the differences, and we got:

$s_d =\frac{\sum_{i=1}^n (d_i -\bar d)^2}{n-1} =11.451$

The 4 step is calculate the statistic given by :

$t=\frac{\bar d -0}{\frac{s_d}{\sqrt{n}}}=\frac{-9.7 -0}{\frac{11.451}{\sqrt{10}}}=-2.679$

The next step is calculate the degrees of freedom given by:

$df=n-1=10-1=9$

Now we can calculate the p value, since we have a left tailed test the p value is given by:

$p_v =P(t_{(9)}$

The p value is higher than the significance level given 0.01, so then we can conclude that we FAIL to reject the null hypothesis. And we can say that the true difference for Best Actresses is not significantly lower than the mean for Best Actors at 1% of significance.

Part b

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The confidence interval for the mean is given by the following formula:

$\bar d \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$ (1)

Since the Confidence is 0.99 or 99%, the value of $\alpha=0.01$ and $\alpha/2 =0.005$ , and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.005,9)".And we see that $t_{\alpha/2}=3.25$

Now we have everything in order to replace into formula (1):

$-9.7-3.25\frac{11.451}{\sqrt{10}}=-21.469$

$-9.7+3.25\frac{11.451}{\sqrt{10}}=2.069$

So on this case the 99% confidence interval would be given by (-21.469;2.069)

Part c

We got the same conclusion as part a, sicne the confidence interval contains the value 0, we FAIL to reject the null hypothesis that the difference between the two means is 0.

8 0

3 years ago

Which graphic organizer would help you visualize important events in the history of the U.S. space program? timeline point-by-po

Sergeu [11.5K]

Timeline would help your the most

6 0

3 years ago

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Hi! This is piecewise equations! Please help if you can :)

Aleonysh [2.5K]

Notice that f(x) is only defined over the intervals 0 ≤ x < 3 and 3 < x ≤ 6; that is, all real numbers x between 0 and 3, and between 3 and 6, but not including 3. Since there is no definition for f(x) at x = 3, the value of f (3) is undefined.

4 0

3 years ago

Solve: (w)(3w+5)=0. Multiple answers, if any, should be separated by commas. Enter only integer(s) or fraction(s).

sp2606 [1]

Answer:

$-\frac{5}{3}$

Step-by-step explanation:

Given expression:

(w)(3w + 5) = 0

Expand the parentheses;

3w² + 5w = 0

3w² = -5w

Divide both sides by w;

3w = - 5

w = $-\frac{5}{3}$

7 0

3 years ago