For example of 100 students the median is 90

Dominic works on the weekends and on vacations from school mowing lawns in his neighborhood. For every lawn he mows, he charges

Ivahew [28]

Answer:

$20\ lawns = \$240$

$9 \ lawns = \$108$

See Attachment for Graph

Step-by-step explanation:

Given

$1\ lawn = \$12$

Solving (a): Number of lawns in $240

$1\ lawn = \$12$

Multiply both sides by 20

$20 * 1\ lawn = \$12 * 20$

$20\ lawns = \$240$

Solving (b): Cost of mowing 9 lawns

$1\ lawn = \$12$

Multiply both sides by 9

$9 * 1\ lawn = \$12 * 9$

$9 \ lawns = \$108$

To create a graph, we need to generate a formula;

If

$1\ lawn = \$12$

$2\ lawn = \$24$

$3\ lawns = \$36$

$x\ lawns = \$12x$

So:

The graph formula is

$y = 12x$

See Attachment for Graph

5 0

3 years ago

John is filling the bathtub that is 18 inches deep. He notices that it takes two minutes to fill the tub with three inches of. W

morpeh [17]

Answer: No he isn't correct because it would take 12 minutes to fill the bathtub up 18 inches if it continued at the same rate

Step-by-step explanation:

3/2 does not equal 18/10

8 0

3 years ago

Suppose the expected tensile strength of type-A steel is 103 ksi and the standard deviation of tensile strength is 7 ksi. For ty

ExtremeBDS [4]

Answer:

a

i So the approximate distribution of $\= X$ is $\mu_{\= X} =103$ and $\sigma_{\= X} = 0.783$

ii So the approximate distribution of $\= Y$ is $\mu_{\= Y} =105$ and $\sigma_{\= Y} = 0.645$

b

the approximate distribution of $\=X - \= Y$ is $E (\= X - \= Y) = -2$ and $\sigma_{\= X - \=Y}=1.029$

Here we can see that the mean of the approximate distribution is negative which tell us that this negative value of the data for $\=X - \= Y$ sample are more and their frequency occurrence is higher than the positive values

c

the value of $P(-1 \le \=X - \= Y \le 1)$ is $= -0.1639$

Step-by-step explanation:

From the question we are given that

The expected tensile strength of the type A steel is $\mu_A = 103 ksi$

The standard deviation of type A steel is $\sigma_A = 7ksi$

The expected tensile strength of the type B steel is $\mu_B = 105\ ksi$

The standard deviation of type B steel is $\sigma_B = 5 \ ksi$

Also the assumptions are

Let $\= X$ be the sample average tensile strength of a random sample of 80 type-A specimens

Here $n_a =80$

Let $\= Y$ be the sample average tensile strength of a random sample of 60 type-B specimens.

Here $n_b = 60$

Let the sampling distribution of the mean be

$\mu _ {\= X} = \mu$

$=103$

Let the sampling distribution of the standard deviation be

$\sigma _{\= X} = \frac{\sigma }{\sqrt{n_a} }$

$= \frac{7}{\sqrt{80} }$

$=0.783$

So What this mean is that the approximate distribution of $\= X$ is $\mu_{\= X} =103$ and $\sigma_{\= X} = 0.783$

For $\= Y$

The sampling distribution of the sample mean is

$\mu_{\= Y} = \mu$

$= 105$

The sampling distribution of the standard deviation is

$\sigma _{\= Y} = \frac{\sigma }{\sqrt{n_b} }$

$= \frac{5}{\sqrt{60} }$

$= 0.645$

So What this mean is that the approximate distribution of $\= Y$ is $\mu_{\= Y} =105$ and $\sigma_{\= Y} = 0.645$

Now to obtain the approximate distribution for $\=X - \= Y$

$E (\= X - \= Y) = E (\= X) - E(\= Y)$

$= \mu_{\= X} - \mu_{\= Y}$

$= 103 -105$

$= -2$

The standard deviation of $\=X - \= Y$ is

$\sigma_{\= X - \=Y} = \sqrt{\sigma_{\= X}^2 - \sigma_{\= Y}^2}$

$= \sqrt{(0.783)^2 + (0.645)^2}$

$=1.029$

Now to find the value of $P(-1 \le \=X - \= Y \le 1)$

Let us assume that $F = \= X - \= Y$

$P(-1 \le F \le 1)$ $= P [\frac{-1 -E (F)}{\sigma_F} \le Z \le \frac{1-E(F)}{\sigma_F} ]$

$= P[\frac{-1-(-2)}{1.029} \le Z \le \frac{1-(-2)}{1.029} ]$

$= P[0.972 \le Z \le 2.95]$

$= P(Z \le 0.972) - P(Z \le 2.95)$

Using the z-table to obtain their z-score

$= 0.8345 - 0.9984$

$= -0.1639$

3 0

3 years ago

A binomial probability experiment is conducted with given parameters. Compute the probability of x successes in the n independen

snow_lady [41]

Answer:

P(X = 4) = 0.1876

Step-by-step explanation:

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

In this question:

$n = 15, p = 0.2$

We want P(X = 4). So

$P(X = 4) = C_{15,4}.(0.2)^{4}.(0.8)^{11} = 0.1876$

7 0

3 years ago

Use the given information to draw a box-and-whisker plot of the data set

drek231 [11]

Answer:

The box-and-whisker plot of this distribution is presented in the attached image to this solution.

Step-by-step explanation:

A box plot gives a visual representation of the distribution of the data, showing where most values lie and those values that greatly differ from the rest, called outliers.

A box and whiskers plot shows 5 major information about the distribution of data. It shows:

- The maximum variable.

- The minimum variable.

- The Median.

- The first quartile.

- The third quartile.

Further info such as the range and Inter quartile range can then be obtained from this 5-number summary.

The elements of the box plot are described thus;

The bottom side of the box represents the first quartile, and the top side, the third quartile. Therefore, the width of the central box represents the inter-quartile range.

The horizontal line inside the box is the median.

The lines extending from the box reach out to the minimum and the maximum values in the data set, as long as these values are not outliers. The ends of the whiskers are marked by two shorter horizontal lines.

Variables in the dataset, higher than Q3+(1.5×IQR) or lower than Q1-(1.5×IQR) are considered outliers and are usually shown using dots above the top whisker or below the bottom whisker.

The required boxplot for this question is given in the attached image to this solution.

The median for the boxplot isn't provided, but it was assumed to be midway between the first and third quartile.

Hope this Helps!!!

7 0

3 years ago