truck that is 10 ft. X 14 ft. X 12 ft. carries boxes that are 1 ft. X 2 ft. X 3 ft. How many of these boxes can this truck hold?

A study was designed to compare the attitudes of two groups of nursing students towards computers. Group 1 had previously taken

Leviafan [203]

Answer:

a. H0:μ1≥μ2

Ha:μ1<μ2

b. t=-3.076

c. Rejection region=[tcalculated<−1.717]

Reject H0

Step-by-step explanation:

a)

As the score for group 1 is lower than group 2,

Null hypothesis: H0:μ1≥μ2

Alternative hypothesis: H1:μ1<μ2

b) t test statistic for equal variances

t=(xbar1-xbar2)-(μ1-μ2)/sqrt[{1/n1+1/n2}*{((n1-1)s1²+(n2-1)s2²)/n1+n2-2}

t=63.3-70.2/sqrt[{1/11+1/13}*{((11-1)3.7²+(13-1)6.6²)/11+13-2}

t=-6.9/sqrt[{0.091+0.077}{136.9+522.72/22}]

t=-3.076

c. α=0.05, df=22

t(0.05,22)=-1.717

The rejection region is t calculated<t critical value

t<-1.717

We can see that the calculated value of t-statistic falls in rejection region and so we reject the null hypothesis at 5% significance level.

7 0

4 years ago

Marissa bought 12.8 pounds of potatoes at $1.35 a pound. How much did Marissa spend on the potatoes?

defon

17.28 dollars is the answer

7 0

4 years ago

Read 2 more answers

Help plsss

LenKa [72]

<h3></h3><h2>⭐ Development:</h2><h3></h3>

Having the expression, let's modify it so it becomes a 2nd degree equation:

$\large {\text {$ \sf \cfrac{1}{2x^2} -2 = 0 $}}$

Now, we will multiply per 2x² both sides of equation...

$\large {\text {$ \sf \cfrac{1}{2x^2}\cdot \:2x^2-2\cdot \:2x^2=0\cdot \:2x^2 $}$

$\searrow$

$\large {\text {$ \sf 1-4x^2=0$}}$

We have to write in standard form...

$\large {\text {$ \sf -4x^2+1 = 0 $}}$

<u />

$\large {\text{$\sf x=\cfrac{-b\pm\sqrt{b^2-4ac}}{2a} \quad\rightarrow\quad x=\cfrac{-0\pm\sqrt{0^2-4\cdot (-4) \cdot1} }{2\cdot (-4) } \:\rightarrow\:\: x=\cfrac{-0^2 \pm4}{2 \cdot(-4)} $}}$

$\huge {\text {$ \sf \downarrow$}}$

$\large {\text {$\sf {\bf x_1} = \cfrac{-0+4}{2\left(-4\right)}= \cfrac{-1}{2} $}}$ $\large {\text {$\sf {\bf x_2 }=\cfrac{-0-4}{2\left(-4\right)} = \cfrac{1}{2} $}}$

At this point, we're going to add the values of x₁ and x₂:

$\large {\boxed {\boxed { \bf x_1 + x_2= -\cfrac{1}{2}+ \cfrac{1}{2} = 0} }}$

$\huge {\text {$ \it Alternative \: A $}}$

8 0

3 years ago

These 3 points are on a parabola defining the edge of a ski:

Tems11 [23]

The equation of the parabola is y = 0.015x^2 + 0.06x + 1

<h3>The linear equations</h3>

A parabola is represented as:

y = ax^2 + bx + c

The points are given as:

(-4,1), (-2,0.94) and (0,1)

When these points are substituted in y = ax^2 + bx + c, we have the following linear equations

16a - 4b + c = 1

4a - 2b + c = 0.94

c = 1

<h3>The matrix that represent the system of equations</h3>

In (a), we have:

16a - 4b + c = 1

4a - 2b + c = 0.94

c = 1

Remove the variables (leaving the coefficients)

a b c

16 -4 1 1

4 -2 1 0.94

0 0 1 1

So, the matrix representation of the system of equations is:

$\left[\begin{array}{ccc}16&-4&1\\4&-2&1\\0&0&1\end{array}\right] \left[\begin{array}{c}1&0.94&1\end{array}\right]$

<h3>The inverse of the coefficient matrix</h3>

Using a graphing calculator, we have the inverse of the coefficient matrix to be

$\left[\begin{array}{ccc}0.125&-0.25&0.125\\0.25&-1&0.75\\0&0&1\end{array}\right]$

<h3>The solution to the system of equations</h3>

Using a graphing calculator, we have the solution to the system of equations to be

a = 0.015

b = 0.06

c = 1

Recall that:

y = ax^2 + bx + c

So, we have:

y = 0.015x^2 + 0.06x + 1

Hence, the equation of the parabola is y = 0.015x^2 + 0.06x + 1