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3 years ago

2. How many solutions does each equation

Mathematics

1 answer:

lakkis [162]3 years ago

7 0

Answer:

2 solution

answer is -c=0 or 0=c

10+3c-2=4c-c+8

3c-4c-c=8-10-2

-c=0

0=c

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The following observations are on stopping distance (ft) of a particular truck at 20 mph under specified experi- mental conditio

zzz [600]

Answer:

see explaination

Step-by-step explanation:

Data : 32.1 , 30.6 , 31.4 , 30.4 , 31.0 , 31.9

Mean X-bar = 31.23

SD = 0.689

Null Hypothesis : Xbar = mu

Alternate Hypothesis : Xbar > mu

z = (Xbar - mu) / (SD/sqrt(n))

= (31.23 - 30 ) /(0.689/sqrt(6))

= 4.372

P-value = ~0

Since P-vale < 0.01 , we will reject null hypothesis.

The data suggest that true average stopping ditance exceeds the maximum.

i) SD = 0.65

mu = 31

z = (Xbar - mu) / (SD/sqrt(n))

= (31.23 - 31) /(0.65/sqrt(6))

= 0.867

P-value = 0.3859 Answer

ii) SD = 0.65

mu = 32

z = (Xbar - mu) / (SD/sqrt(n))

= (31.23 - 32) /(0.65/sqrt(6))

= -2.9

P-value = 0.0037 Answer

i) SD = 0.8

mu = 31

z = (Xbar - mu) / (SD/sqrt(n))

= (31.23 - 31) /(0.8/sqrt(6))

= 0.704

P-value = 0.4814 Answer

ii) SD = 0.8

mu = 32

z = (Xbar - mu) / (SD/sqrt(n))

= (31.23 - 32) /(0.8/sqrt(6))

= -2.357

P-value = 0.0184 Answer

The probabilities obtained in part c are comparatively higher than that of part b.

i) For alpha =0.01

z = (Xbar - mu) / (SD/sqrt(n))

=> -2.32 = (31.23 - 31) /(0.65/sqrt(n))

=> (0.65/sqrt(n)) = (31.23 - 31)/-2.32

=> sqrt(n) = (0.65*(-2.32)) / (31.23 - 31)

=> n = 43 Answer

ii) For beta =0.10

z = (Xbar - mu) / (SD/sqrt(n))

=> -1.28 = (31.23 - 31) /(0.65/sqrt(n))

=> (0.65/sqrt(n)) = (31.23 - 31)/-1.28

=> sqrt(n) = (0.65*(-1.28)) / (31.23 - 31)

=> n = 13 Answer

4 0

3 years ago

A number added to 12 and then subtracted by 18 equals 26. What is the number

tiny-mole [99]

The answer would be 32. take your 26 and add 18 then subtract 12. then rework it forward if you don't understand.

4 0

3 years ago

Read 2 more answers

Three security cameras were mounted at the corners of a triangles parking lot. Camera 1 was 110 ft from camera 2, which was 137

Nata [24]

Answer:

<em>Camera 2nd has to cover the maximum angle, i.e. </em> $78.70^\circ$ .

Step-by-step explanation:

Please have a look at the triangular park represented as a triangle $\triangle ABC$ with sides

a = 110 ft

b = 158 ft

c = 137 ft

1st camera is located at point C, 2nd camera at point B and 3rd camera at point A respectively.

We can use law of cosines here, to find out the angles $\angle A, \angle B, \angle C$

As per Law of cosine:

$cos C = \dfrac{a^{2}+b^2-c^2 }{2ab}\\cos B = \dfrac{a^{2}+c^2-b^2 }{2ac}\\cos A = \dfrac{b^{2}+c^2-a^2 }{2bc}$

Putting the values of a,b and c to find out angles $\angle A, \angle B, \angle C$ .

$cos C = \dfrac{110^{2}+158^2-137^2 }{2\times 110 \times 158}\\\Rightarrow cos C = \dfrac{12100+24964-18769 }{24760}\\\Rightarrow cos C =0.526\\\Rightarrow C = 58.24^\circ$

$cos B = \dfrac{110^{2}+137^2-158^2 }{2\times 110 \times 137}\\\Rightarrow cos B = \dfrac{12100+18769 -24964}{30140}\\\Rightarrow cos B = \dfrac{5905}{30140}\\\Rightarrow cos B =0.196\\\Rightarrow B = 78.70^\circ$

$cos A = \dfrac{158^{2}+137^2-110^2 }{2\times 158 \times 137}\\\Rightarrow cos A = \dfrac{24964+18769-12100}{43292}\\\Rightarrow cos A = \dfrac{31633}{43292}\\\Rightarrow cos A = 0.731\\\Rightarrow A = 43.05^\circ$

<em>Camera 2nd has to cover the maximum angle</em>, i.e. $78.70^\circ$ .

6 0

4 years ago

An instructor at a major research university occasionally teaches summer session and notices that that there are often students

Mademuasel [1]

Answer:

Check the explanations

Step-by-step explanation:

According to given information, an instructor at a major research university occasionally teaches summer session and notices that that there are often students repeating the class. On the first day of class, she counts 105 students enrolled, of which 19 are repeating the class. The university enrolls 15,000 students.

Therefore the number of observation n 105 and enrolled x=19

1. An estimate of the population proportion repeating the class is given by:

c) 0.181.

Explanation:

19 --0.18090,181 n 105

2. The instructor wishes to estimate the proportion of students across campus who repeat a course during summer sessions and decides to do so on the basis of this class. Would you advise the instructor against it and why:

d) Yes, because this class is not a random sample of students.

3. The standard error for the estimated sample proportion is given by:

c) 0.0376.

Explanation:

SE $(\hat{p})=\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$

0.181 × (1-0.181) 105

0,0376 0.0376

4. A 95% confidence interval is given by:

c) 0.107, 0.255.

Explanation:

In order to determine the 95% confidence interval we follow the following step:

Where the z value is determined from the standard normal table as ~ 1.96

$\hat{p}\pm \left [z\times SE(\hat{p}) \right ]$

0.181 ± (1.96 × 0.0376)

0.181 ± 0.0737

Therefore the lower confidence interval is

LCI= 0.181- 0.0737

LCI= 0.107

Therefore the upper confidence interval is

UCI = 0.181 + 0.0737

UCI0.255

Therefore 95% confidence interval is

$\left (0.107,0.255 \right )$

5. She hypothesizes that, in general, 10% of students repeat a course. The hypotheses to be tested are:

d) $0 :P = 0.10 and H_{a}:p\neq 0.10$

Explanation:

She hypothesizes that, in general, 10% of students repeat a course. The hypotheses to be tested are defined on the basis of observation,

Null hypothesis as:

$H_{0}$ :p= 0.10

and alternative hypothesis as:

$H_{a}:p\neq 0.10$

6. The test statistic for this hypothesis is given by:

c) 2.765.

Explanation:

In order to determine the z test statistics as:

Z= $\frac{\hat{p}-p_{0}}{\sqrt{\frac{p_{0}(1-p_{0})}{n}}}$

0.181 - 0.10 0.10x (1-0.10)

=2.765

7. The P value for this test is:

c) 0.01>P>0.005 .

Explanation:

P value is calculated as:

P(Z > 2.765) 0.002845 for one tail test and

P(Z > 2.765) 0.005692 for two tail test.

8. Based on the p-value found:

b) we have strong evidence that the proportion of students repeating a class during summer sessions is not 10%.

Explanation:

As the z observed is more than the tabulated z value at 95% as:

$Z_{observed}=2.765> Z_{tabulated}=1.96$

and also P value is less than the $\alpha =1-0.95=0.05$

$P(Z\geq 2.765)=0.005692< \alpha =0.05$

Therefore we accept the alternative hypothesis and we may conclude that we have strong evidence that the proportion of students repeating a class during summer sessions is not 10%.

7 0

3 years ago

Vito avido is a sales representative for a company that sells roofing material. he receives a graduated commission of 5 percent

nexus9112 [7]