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Igoryamba
3 years ago
15

Pls help thanks alot in advance ​

Mathematics
2 answers:
riadik2000 [5.3K]3 years ago
7 0

Answer:A=true B.=true    C= false  D=true

Step-by-step explanation:

put them in to an algebra calculator

Rashid [163]3 years ago
4 0

Answer:

A. True

B. True

C. False

D. True

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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
Find the indefinite integral. (Note: Solve by the simplest method—not all require integration by parts. Use C for the constant o
umka2103 [35]

Answer:

∫▒〖arctan⁡(x).1 dx=arctan⁡(x).x〗-1/2  ln⁡(1+x^2 )+C

Step-by-step explanation:

∫▒〖1st .2nd dx=1st∫▒〖2nd dx〗-∫▒〖(derivative of 1st) dx∫▒〖2nd dx〗〗〗

Let 1st=arctan⁡(x)

And 2nd=1

∫▒〖arctan⁡(x).1 dx=arctan⁡(x) ∫▒〖1 dx〗-∫▒〖(derivative of arctan(x))dx∫▒〖1 dx〗〗〗

As we know that  

derivative of arctan(x)=1/(1+x^2 )

∫▒〖1 dx〗=x

So  

∫▒〖arctan⁡(x).1 dx=arctan⁡(x).x〗-∫▒〖(1/(1+x^2 ))dx.x〗…………Eq1

Let’s solve ∫▒(1/(1+x^2 ))dx by substitution now  

Let 1+x^2=u

du=2xdx

Multiply and divide ∫▒〖(1/(1+x^2 ))dx.x〗 by 2 we get

1/2 ∫▒〖(2/(1+x^2 ))dx.x〗=1/2 ∫▒(2xdx/u)  

1/2 ∫▒(2xdx/u) =1/2 ∫▒(du/u)  

1/2 ∫▒(2xdx/u) =1/2  ln⁡(u)+C

1/2 ∫▒(2xdx/u) =1/2  ln⁡(1+x^2 )+C

Putting values in Eq1 we get

∫▒〖arctan⁡(x).1 dx=arctan⁡(x).x〗-1/2  ln⁡(1+x^2 )+C  (required soultion)

3 0
3 years ago
Read 2 more answers
Based on FAA estimates the average age of the fleets of the 10 largest U.S. commercial passenger carriers is 13.4 years with a s
Aliun [14]

Answer : 0.0129

Step-by-step explanation:

Given : Based on FAA estimates the average age of the fleets of the 10 largest U.S. commercial passenger carriers is \mu=13.4 years and standard deviation is \sigma=1.7 years.

Sample size : n=40

Let X be the random variable that represents the age of fleets.

We assume that the ages of the fleets of the 10 largest U.S. commercial passenger carriers are normally distributed.

For z-score,

z=\dfrac{x-\mu}{\dfrac{\sigma}{\sqrt{n}}}

For x=14

z=\dfrac{14-13.4}{\dfrac{1.7}{\sqrt{40}}}\approx2.23

By using the standard normal distribution table , the probability that the average age of these 40 airplanes is at least 14 years old will be :-

P(x\geq 14)=P(z\geq2.23)\\\\=1-P(z

Hence, the required probability = 0.0129

6 0
3 years ago
25 POINTS!!!!!! PLEEASEEEE HELP!!!!!!!!!!!!!!NO LINKS OR SPAMS I already solved half of it pls help on part 4 and down.
vichka [17]

Answer:

Death Valley California

C(t) = -0.30 (0)² + 40

C(t) = 40

C(t) = -0.30 (12)² + 40t – 12)2 + 40

65f

7 0
3 years ago
Please help I will appreciate it
LiRa [457]

Answer:

For part A the value of m is 2/5

Step-by-step explanation:

When a system equation has no solution it means a equation is parallel and it doesn't intersect. And when a line is parallel to one another it has the same slope.

8 0
3 years ago
Read 2 more answers
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