2.6 X 16 fourth grade answers

Snowcat [4.5K]

X=6 is the answer for the problem

6 0

3 years ago

A student believes that less than 50% of students at his college receive financial aid. A random sample of 120 students was take

marusya05 [52]

Answer:

p= 0.9995 and we can conclude that students receiving aid is more than 50% according to the sample results in 2% significance level.

Step-by-step explanation:

$H_{0}$ : less than or equal 50% of students at his college receive financial aid

$H_{a}$ : more than 50% of students receive financial aid.

According to the nul hypothesis we assume a normal distribution with proportion 50%.

z-score of sample proportion can be calculated using the formula:

z= $\frac{X-M}{\frac{s}{\sqrt{N} } }$ where

X is the sample proportion (0.65)
M is the null hypothesis proportion (0.5)
s is the standard deviation of the sample ( $\sqrt{0.5*(1-0.5)}$ )
N is the sample size (120)

then z= $\frac{0.65-0.50}{\frac{\sqrt{0.25}}{\sqrt{120} } }$ ≈ 3,29

thus p= 0.9995 and since p<0.02 (2%), we reject the null hypothesis.

4 0

3 years ago

Santa Claus is assigning elves to work an eight-hour shift making toy trucks. Apprentice earn five candy canes per hour but can

dimaraw [331]

Answer:

(a) 5 senior, 4 apprentice

(b) 368 per shift

(c) 7.5 senior, 0 apprentice

Step-by-step explanation:

The problem can be described by two inequalities. On describes the limit on the number of elves in the shop; the other describes the limit on the total payroll. Let x and y represent the number of apprentice and senior elves, respectively. Then the inequalities for the first scenario are ...

x + y ≤ 9 . . . . . . . . . . . . total number of elves in the shop

5x +8y ≤ 480/8 . . . . . . candy canes per hour paid to elves

These two inequalities are graphed in the first attachment. They describe a solution space with vertices at ...

(x, y) = (0, 7.5), (4, 5), (9, 0)

__

(a) Santa wants to maximize the output of trucks, so wants to maximize the function t = 4x +6y.

At the vertices of the solution space, the values of this function are ...

t(0, 7.5) = 45

t(4, 5) = 46

t(9, 0) = 36

Output of trucks is maximized by a workforce of 4 apprentice elves and 5 senior elves.

__

(b) The above calculations show 46 trucks per hour can be made, so ...

46×8 = 368 . . . trucks in an 8-hour shift

__

(c) The new demands change the inequalities to ...

x + y ≤ 8 . . . . . . number of workers

7x +8y ≤ 60 . . . total wages (per hour)

The vertices of the feasible region for these condtions are ...

(x, y) = (0, 7.5), (4, 4), (8, 0)

From above, we know the truck output will be maximized at the vertex (x, y) = (0, 7.5). However, we know we cannot have 7.5 senior elves working in the shop. We can have 7 or 8 elves working.

If the workforce must remain constant, truck output is maximized by a workforce of 7 senior elves.

If the workforce can vary through the shift, truck output is maximized by adding one more senior elf in the shop for half a shift.

Santa should assign 7 senior elves for the entire shift, and 8 senior elves (one more) for half a shift.

_____

<em>Comment on apprentice elf wages</em>

At 5 candy canes for 4 trucks, apprentice elves produced trucks for a cost of 1.25 candy canes per truck. At 8 candy canes for 6 trucks, senior elves produced trucks for a cost of about 1.33 candy canes per truck. The reason for employing senior elves in the first scenario is that their productivity is 1.5 times that of apprentice elves while their cost per truck is about 1.07 times that of apprentice elves.

After the apprentice elves wages were increased, their cost per truck is 1.75 candy canes per truck, but their productivity hasn't changed. They have essentially priced themselves out of a job, because they are not competitive with senior elves.

5 0

3 years ago

Luis is packing his bags for his vacation he has 9 unique toy animals but only 5 fit in his bag how many different groups of 5 t

Eva8 [605]

Answer:

126 groups

Step-by-step explanation:

we know that

This is a combination problem, because the order doesn't matter. He has 9 toys and wants to know the number of different combinations of 5 he can take.

The formula for the number of possible combinations of r objects from a set of n objects is given by

$^nC_r=\frac{ {n!}}{r!(n-r)!}$

In this problem we have

$n=9, r=5$

substitute

$^9C_5=\frac{ {9!}}{5!(9-5)!}$

$^9C_5=\frac{ {9!}}{5!(4)!}$

$^9C_5=\frac{ {(9)(8)(7)(6)5!}}{5!(4)!}$

$^9C_5=\frac{ {9)(8)(7)(6)}}{(4*3*2*1)}$

$^9C_5=126$

7 0

3 years ago

Cheryl collected data for her mathematics project. She noted that the data set was approximately normal.

nadya68 [22]

Answer:

$X_(r) >> X_(n)$

The mean for this case would increase since is defined as:

$\bar X= \frac{\sum_{i=1}^n X_i}{n}$

The interquartile range would not change since the definition for the IQR is $IQR =Q_3 -Q_1$ and the quartiles are the same.

The standard deviation would not remain the same since by definition is:

$s = \sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}}$

And since we change the largest value the deviation would increase considerably.

And for the last option is not always true since if we select a value so much higher then the distribution would be skewed to the right.

So the best option for this case is:

Mean would increase.

Step-by-step explanation:

For this case we assume that we have a random sample given $X_(1), X_(2) ,..., X_(n)$ and for each observation $X_i \sim N(\mu, \sigma)$ since the problem states that the data is approximately normal.

Let's assume that the largest value on this sample is $X_(n)$ and for this case we are going to replace this value by another one extremely higher so we satisfy this condition:

$X_(r) >> X_(n)$

The mean for this case would increase since is defined as:

$\bar X= \frac{\sum_{i=1}^n X_i}{n}$

The interquartile range would not change since the definition for the IQR is $IQR =Q_3 -Q_1$ and the quartiles are the same.

The standard deviation would not remain the same since by definition is:

$s = \sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}}$

And since we change the largest value the deviation would increase considerably.

And for the last option is not always true since if we select a value so much higher then the distribution would be skewed to the right.

So the best option for this case is:

Mean would increase.

6 0

3 years ago