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3 years ago

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Muna has 23 cars available to rent. Let c be the number of cars she would have left after r renting of them. Write an equation r

elating c to r . Then use this equation to find the number of cars she would have left after renting 14 of them.

1 answer:

BigorU [14]3 years ago

4 0

Answer:

Step-by-step explanation:

um

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Consider the following hypothesis test.H0:μ1−μ2=0 Ha:μ1−μ2≠0The following results are for two independent samples taken from the

Julli [10]

Answer:

a) $z =\frac{104-106}{\sqrt{\frac{8.4^2}{80} +\frac{7.6^2}{70}}}= -1.53$

b) $p_v =2*P(z$

c) Since the p value is higher than the significance level provided we have enogh evidence to FAIL to reject the null hypothesis and we can't conclude that the true means are different at 5% of significance

Step-by-step explanation:

Information given

$\bar X_{1}= 104$ represent the mean for 1

$\bar X_{2}= 106$ represent the mean for 2

$\sigma_{1}= 8.4$ represent the population standard deviation for 1

$\sigma_{2}= 7.6$ represent the population standard deviation for 2

$n_{1}=80$ sample size for the group 1

$n_{2}=70$ sample size for the group 2

z would represent the statistic

Hypothesis to test

We want to check if the two means for this case are equal or not, the system of hypothesis would be:

H0: $\mu_{1}=\mu_{2}$

H1: $\mu_{1} \neq \mu_{2}$

The statistic would be given by:

$z =\frac{\bar X_1-\bar X_2}{\sqrt{\frac{\sigma^2_1^2}{n_1} +\frac{\sigma^2_2^2}{n_2}}}=$ (1)

Part a

Replacing we got:

$z =\frac{104-106}{\sqrt{\frac{8.4^2}{80} +\frac{7.6^2}{70}}}= -1.53$

Part b

The p value would be given by this probability:

$p_v =2*P(z$

Part c

Since the p value is higher than the significance level provided we have enogh evidence to FAIL to reject the null hypothesis and we can't conclude that the true means are different at 5% of significance

6 0

3 years ago

Find the area of the triangle ABC with the coordinates of A(10, 15) B(15, 15) C(30, 9).

lions [1.4K]

Check the picture below. so, that'd be the triangle's sides hmmm so let's use Heron's Area formula for it.

$~\hfill \stackrel{\textit{\large distance between 2 points}}{d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2}}~\hfill~ \\\\[-0.35em] ~\dotfill\\\\ (\stackrel{x_1}{10}~,~\stackrel{y_1}{5})\qquad (\stackrel{x_2}{15}~,~\stackrel{y_2}{15}) ~\hfill a=\sqrt{[ 15- 10]^2 + [ 15- 5]^2} \\\\\\ ~\hfill \boxed{a=\sqrt{125}} \\\\\\ (\stackrel{x_1}{15}~,~\stackrel{y_1}{15})\qquad (\stackrel{x_2}{30}~,~\stackrel{y_2}{9}) ~\hfill b=\sqrt{[ 30- 15]^2 + [ 9- 15]^2} \\\\\\ ~\hfill \boxed{b=\sqrt{261}}$

$(\stackrel{x_1}{30}~,~\stackrel{y_1}{9})\qquad (\stackrel{x_2}{10}~,~\stackrel{y_2}{5}) ~\hfill c=\sqrt{[ 10- 30]^2 + [ 5- 9]^2} \\\\\\ ~\hfill \boxed{c=\sqrt{416}} \\\\[-0.35em] ~\dotfill$

$\qquad \textit{Heron's area formula} \\\\ A=\sqrt{s(s-a)(s-b)(s-c)}\qquad \begin{cases} s=\frac{a+b+c}{2}\\[-0.5em] \hrulefill\\ a=\sqrt{125}\\ b=\sqrt{261}\\ c=\sqrt{416}\\ s\approx 23.87 \end{cases} \\\\\\ A\approx\sqrt{23.87(23.87-\sqrt{125})(23.87-\sqrt{261})(23.87-\sqrt{416})}\implies \boxed{A\approx 90}$

6 0

2 years ago

How should you multiply fractions?

Verizon [17]

Answer:

DDDDD...............................

5 0

2 years ago

Read 2 more answers

Please help me there can only be one answer

12345 [234]

Answer:-
I think it’s the last option where x and y both are zero.
2x + 2y = 0
Here x = 0 and y = 0.
Hope it helps if u hv any questions feel free to ask...hv a grt day :)

4 0

3 years ago

Alex is a writer who writes poems and short stories. For an upcoming writer's workshop Alex wants to write some new works. He ne

pashok25 [27]

Answer:

The maximum number of works that he can write while staying in his time budget is 24.

21 poems and 3 short stories

Step-by-step explanation:

In order to solve this problem we must first determine what our variables are. In this case it's the number of poems and short stories he can write.

p = # of poems

s = # of short stories

Next, we must build our objective function which will represent the total number of works he can write.

N=p+s

where N is the number of works.

Next, we must write the constrains based on the information provided by the problem.

The problem tells us that it takes him 30 hours to write a poem and 70 hours to write a short story and that he has 840 hours available to write them, so that constrain will be the following:

$30p+70s \leq 840$

it also tells us that he wants to write at least 4 poems and 3 short stories so there we have our other two constrains.

$p \geq 4$

$s \geq 3$

once we got our constrains we can go ahead and graph them to see how they will behave. (See attached picture)

In the graph p is the horizontal axis and s is the vertical axis.

On the graph we can see a polygon that is formed by the restriction. The vertices of the polygon will represent the optimal conditions for this linear programming problem. There are three optimal solutions there, so we need to test them to see which will return the greatest number of works he can write while keeping the given conditions.

Option 1:

4 poems and 3 short stories

N=4+3

N= 7 works

Option 2:

4 poems and 10 short stories

N=4+10

N=14 works

Option 3:

21 poems and 3 short stories

N=21+3

N=24 works

So the optimal solution will be given by option 3 with 21 poems and 3 short stories.

5 0

3 years ago