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3 years ago

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A car dealership has three locations, all about the same size, within the city. The graph shows last month’s sales figures for o

ne of the locations. Which statement is best supported by the information in the graph? on ttm!

1 answer:

postnew [5]3 years ago

6 0

Answer:

Truck sales across all stores were likely 60

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Lightbulbs of a certain type are advertised as having an average lifetime of 750 hours. The price of these bulbs is very favorab

Katyanochek1 [597]

Answer:

(a) Customer will not purchase the light bulbs at significance level of 0.05

(b) Customer will purchase the light bulbs at significance level of 0.01 .

Step-by-step explanation:

We are given that Light bulbs of a certain type are advertised as having an average lifetime of 750 hours. A random sample of 50 bulbs was selected, and the following information obtained:

Average lifetime = 738.44 hours and a standard deviation of lifetimes = 38.2 hours.

Let Null hypothesis, $H_0$ : $\mu$ = 750 {means that the true average lifetime is same as what is advertised}

Alternate Hypothesis, $H_1$ : $\mu$ < 750 {means that the true average lifetime is smaller than what is advertised}

Now, the test statistics is given by;

T.S. = $\frac{Xbar-\mu}{\frac{s}{\sqrt{n} } }$ ~ $t_n_-_1$

where, X bar = sample mean = 738.44 hours

s = sample standard deviation = 38.2 hours

n = sample size = 50

So, test statistics = $\frac{738.44-750}{\frac{38.2}{\sqrt{50} } }$ ~ $t_4_9$

= -2.14

(a) Now, at 5% significance level, t table gives critical value of -1.6768 at 49 degree of freedom. Since our test statistics is less than the critical value of t, so which means our test statistics will lie in the rejection region and we have sufficient evidence to reject null hypothesis.

Therefore, we conclude that the true average lifetime is smaller than what is advertised and so consumer will not purchase the light bulbs.

(b) Now, at 1% significance level, t table gives critical value of -2.405 at 49 degree of freedom. Since our test statistics is higher than the critical value of t, so which means our test statistics will not lie in the rejection region and we have insufficient evidence to reject null hypothesis.

Therefore, we conclude that the true average lifetime is same as what it has been advertised and so consumer will purchase the light bulbs.

6 0

3 years ago

A car passes a landmark traveling at a constant rate of 55 kilometers per hour. One hour later, a second car passes the same lan

yarga [219]

Recall your d = rt, distance = rate * time

one car is going at 55kph, at say hmmm "t" time
the second car zooms by at 75kph, an hour later, thus " t + 1 "

notice, overtaking the 1st car, simply means, the second car comes from behind approaches the first car from behind and "gets there", at that very second, both cars have travelled a distance, say "d"

if the first car travelled "d" kilometers
for the second car to "get there" has to had travelled "d" distance as well

$\bf \begin{array}{lccclll}
&distance&rate&time\\
&-----&-----&-----\\
\textit{first car}&d&55&t\\
\textit{second car}&d&75&t+1
\end{array}
\\\\\\
\begin{cases}
d=55t\\
d=75(t+1)\\
-------\\
55t=75(t+1)
\end{cases}$

solve for "t"

7 0

3 years ago

Read 2 more answers

Given the center (h,k) = (5,0) and radius r=3, write and equation for the circle in standard form

Len [333]

Answer

B) $(x-5)^{2} + y^{2} = 9$

Step by step explanation

The standard form of circle when center(h, k) and radius r, given is

(x - h)^2 + (y - k)^2 = r^2

Given: (h, k) = (5, 0) and r = 3

Now plug in these values in the standard form, we get

(x - 5)^2 + (y - 0)^2 = 3^2

$(x- 5)^{2} + y^{2} = 9$

Therefore, the answer is B) $(x-5)^{2} + y^{2} = 9$

Thank you :)

3 0

2 years ago

Read 2 more answers

2 of the people on a train have

Keith_Richards [23]

Answer:

10

Step-by-step explanation:

5 0

2 years ago

Which is the best approximation to a solution of the equation e^x = 2x + 3?

TiliK225 [7]

The correct question is
Which is the best approximation to a solution of the equation
e^(2x) = 2e^{x) + 3?

we have that

e^(2x) = 2e^{x) + 3-----------> e^(2x)- 2e^{x) - 3=0
the term
e^(2x)- 2e^{x)----------> (e^x)²-2e^(x)*(1)+1²-1²------> (e^x-1)²-1

then
e^(2x)- 2e^{x) - 3=0--------> (e^x-1)²-1-3=0------> (e^x-1)²=4
(e^x-1)=2--------> e^x=3
x*ln(e)=ln(3)---------> x=ln(3)
ln(3)=1.10
hence
x=1.10

the answer is x=1.10

3 0

3 years ago