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GREYUIT [131]
9 months ago
8

Carter and Isabella run a carwash business. They split the revenue 50/50. They are deciding on if they should join forces with C

arla's carwash.If they join with Carla, the combined carwash will make $2400 more per month, but they'll have to split revenue equally (33.3/33.3/33.3). On an average month, Carter and Isabella's carwash makes $6000. How much more or less would Carter or Isabella individually make if they merged with Carla's carwash? This question is required. *A) $200 lessB) $400 moreC) $200 moreD) $400 less
Mathematics
1 answer:
Ivahew [28]9 months ago
6 0

ANSWER

A) $200 less

EXPLANATION

To know how much more or less Isabelle or Carter will make if they merged with Carla's carwash, we have to find the difference between how much each of them makes now and how much they will make if they join forces with Carla.

Right now, Isabella and Carter make $6,000 a month. To find how much each of them makes, divide by 2 since they share the money 50/50 i.e. equally:

\begin{gathered} \frac{6000}{2} \\ \Rightarrow\$3,000 \end{gathered}

They make $3,000 each.

If they join forces with Carla, they will make $2,400 more per month. That is:

\begin{gathered} \$6,000+\$2,400 \\ \Rightarrow\$8,400 \end{gathered}

To find how much each of them will make, divide by 3 since they split it equally i.e. 33.3/33.3/33.3:

\begin{gathered} \frac{8,400}{3} \\ \Rightarrow\$2,800 \end{gathered}

Now, we have to find the difference between the values:

\begin{gathered} \$2,800-\$3,000 \\ \Rightarrow-\$200 \end{gathered}

As we can see, they will lose $200 per month if they join forces with Carla.

Hence, they will individually make $200 less if they merged with Carla's carwash.

The answer is option A.

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Use a t-distribution to answer this question. Assume the samples are random samples from distributions that are reasonably norma
Nataliya [291]

Answer:

The degrees of freedom is 11.

The proportion in a t-distribution less than -1.4 is 0.095.

Step-by-step explanation:

The complete question is:

Use a t-distribution to answer this question. Assume the samples are random samples from distributions that are reasonably normally distributed, and that a t-statistic will be used for inference about the difference in sample means. State the degrees of freedom used. Find the proportion in a t-distribution less than -1.4  if the samples have sizes 1 = 12 and n 2 = 12 . Enter the exact answer for the degrees of freedom and round your answer for the area to three decimal places. degrees of freedom = Enter your answer; degrees of freedom proportion = Enter your answer; proportion

Solution:

The information provided is:

n_{1}=n_{2}=12\\t-stat=-1.4

Compute the degrees of freedom as follows:

\text{df}=\text{Min}.(n_{1}-1,\ n_{2}-1)

   =\text{Min}.(12-1,\ 12-1)\\\\=\text{Min}.(11,\ 11)\\\\=11

Thus, the degrees of freedom is 11.

Compute the proportion in a t-distribution less than -1.4 as follows:

P(t_{df}

                      =P(t_{11}>1.4)\\\\=0.095

*Use a <em>t</em>-table.

Thus, the proportion in a t-distribution less than -1.4 is 0.095.

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3 years ago
Solve the equation by graphing. If exact roots cannot be found, state the consecutive integers between which the roots are locat
zavuch27 [327]

Answer:

The equation contains exact roots at x = -4 and x = -1.

See attached image for the graph.

Step-by-step explanation:

We start by noticing that the expression on the left of the equal sign is a quadratic with leading term x^2, which means that its graph shows branches going up. Therefore:

1) if its vertex is ON the x axis, there would be one solution (root) to the equation.

2) if its vertex is below the x-axis, it is forced to cross it at two locations, giving then two real solutions (roots) to the equation.

3) if its vertex is above the x-axis, it will not have real solutions (roots) but only non-real ones.

So we proceed to examine the vertex's location, which is also a great way to decide on which set of points to use in order to plot its graph efficiently:

We recall that the x-position of the vertex for a quadratic function of the form f(x)=ax^2+bx+c is given by the expression: x_v=\frac{-b}{2a}

Since in our case a=1 and b=5, we get that the x-position of the vertex is: x_v=\frac{-b}{2a} \\x_v=\frac{-5}{2(1)}\\x_v=-\frac{5}{2}

Now we can find the y-value of the vertex by evaluating this quadratic expression for x = -5/2:

y_v=f(-\frac{5}{2})\\y_v=(-\frac{5}{2} )^2+5(-\frac{5}{2} )+4\\y_v=\frac{25}{4} -\frac{25}{2} +4\\\\y_v=\frac{25}{4} -\frac{50}{4}+\frac{16}{4} \\y_v=-\frac{9}{4}

This is a negative value, which points us to the case in which there must be two real solutions to the equation (two x-axis crossings of the parabola's branches).

We can now continue plotting different parabola's points, by selecting x-values to the right and to the left of the x_v=-\frac{5}{2}. Like for example x = -2 and x = -1 (moving towards the right) , and x = -3 and x = -4 (moving towards the left.

When evaluating the function at these points, we notice that two of them render zero (which indicates they are the actual roots of the equation):

f(-1) = (-1)^2+5(-1)+4= 1-5+4 = 0\\f(-4)=(-4)^2+5(-4)_4=16-20+4=0

The actual graph we can complete with this info is shown in the image attached, where the actual roots (x-axis crossings) are pictured in red.

Then, the two roots are: x = -1 and x = -4.

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3 years ago
A certain virus infects one in every 300 people. A test used to detect the virus in a person is positive 90​% of the time when t
garik1379 [7]

Answer:

a)  P[A/B] = 0,019     or     P[A/B] = 1,9 %

b)  P[A- /B-] = 0,9996       or    P[A- /B-] = 99,96 %

Step-by-step explanation:

Bayes Theorem :

P[A/B]  =  P(A) * P[B/A] / P(B)

The branches of events are as follows

Condition 1        real infection     1/300        and     not infection  299/300

Then

1.-    1/300      299/300

When the test is done   (virus present)  0,9 (+)    0,15 (-)

2.-   299/300

When the test is done  ( no virus )   0,15  (+)     0,85 (-)

Then:

P(A) = event person infected          P(B)  =  person test positive

a) P[A/B]  = P(A) * P[B/A] / P(B)

where   P(A)  = 1/300  =   0,0033   P[B/A] = 0,9    

Then P(A) * P[B/A] =  0,0033*0,9  =  0,00297

P(B)   is    ( 1/300 )*0,9  +  (299/300)*0,15

P(B) = 0,0033*0,9 + 0,9966*0,15    ⇒  P(B) = 0,1524

Finally

P[A/B] =  0,00297 /0,1524

P[A/B] = 0,019     or     P[A/B] = 1,9 %

b) Following sames steps:

P[A- /B-] = (299/300) * 0,85  / (299/300) * 0,85 + (1/300 * 0,1)

P[A- /B-] = 0,8471 /0,8474

P[A- /B-] = 0,9996       or    P[A- /B-] = 99,96 %

6 0
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