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DiKsa [7]
3 years ago
13

Determine the number of solutions for the equation shown below

Mathematics
2 answers:
Mila [183]3 years ago
5 0
The answer would be B
polet [3.4K]3 years ago
5 0
It should be b because when solving for x, you have to get x on one side of the equation so when subtracting 2x, it’ll equal 0 leaving the question unsolvable
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Whats 1/10 of 6,000<br><br><br>​
Katen [24]

Answer:

600

Step-by-step explanation:

(1 / 10) * 6000

That's it!

6 0
3 years ago
Please help me ----------------
Brut [27]

Answer:

C = 43.98 in

Step-by-step explanation:

Radius = 7

C / d = pi

C = pi * d

C = pi * 14

7 0
3 years ago
Graph XY with endpoints X(5,−2) and Y(3,−3) and its image after a reflection in the x-axis and then a rotation of 270 degrees ab
DerKrebs [107]

Answer:

  X''(2, -5), Y''(3, -3)

Step-by-step explanation:

You know that reflection in the x-axis changes the sign of the y-coordinate. Points that used to be above the axis are now below by the same amount, and vice versa.

Rotation counterclockwise by 270° is the same as clockwise rotation by 90°. That maps the coordinates like this:

  (x, y) ⇒ (y, -x)

The two transformations together give you ...

  (x, y) ⇒ (x, -y) ⇒ (-y, -x) . . . . . . . . equivalent to reflection across y=-x.

Using this mapping, we have ...

  X(5, -2) ⇒ X''(2, -5)

  Y(3, -3) ⇒ Y''(3, -3) . . . . . . on the equivalent line of reflection, so invariant

_____

The attachment shows the original segment in red, the reflected segment in purple, and the rotated segment in blue. The equivalent line of reflection is shown as a dashed green line.

8 0
3 years ago
A random sample of 25 ACME employees showed the average number of vacation days taken during the year is 18.3 days with a standa
Norma-Jean [14]

Answer:

a) Null hypothesis:\mu \leq 15  

Alternative hypothesis:\mu > 15  

b) df=n-1=25-1=24  

For this case the p value is given p_v = 0.0392

If we compare the p value and the significance level given \alpha=0.05 we see that p_v so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the true mean is higher than 15 at 5% of signficance.  

c) Type I error, also known as a “false positive” is the error of rejecting a null  hypothesis when it is actually true. Can be interpreted as the error of no reject an  alternative hypothesis when the results can be  attributed not to the reality.

So for this case a type I of error would be reject the hypothesis that the true mean is less or equal than 15 and is actually true.

Step-by-step explanation:

Data given and notation  

\bar X=18.3 represent the sample mean

s=3.72 represent the sample standard deviation

n=25 sample size  

\mu_o =15 represent the value that we want to test

\alpha=0.05 represent the significance level for the hypothesis test.  

t would represent the statistic (variable of interest)  

p_v represent the p value for the test (variable of interest)  

Part a: State the null and alternative hypotheses.  

We need to conduct a hypothesis in order to check if the true mean for vacation days is higher than 15, the system of hypothesis would be:  

Null hypothesis:\mu \leq 15  

Alternative hypothesis:\mu > 15  

If we analyze the size for the sample is < 30 and we don't know the population deviation so is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:  

t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}  (1)  

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".  

Part b: P-value  and conclusion

The first step is calculate the degrees of freedom, on this case:  

df=n-1=25-1=24  

For this case the p value is given p_v = 0.0392

Conclusion  

If we compare the p value and the significance level given \alpha=0.05 we see that p_v so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the true mean is higher than 15 at 5% of signficance.  

Part c

Type I error, also known as a “false positive” is the error of rejecting a null  hypothesis when it is actually true. Can be interpreted as the error of no reject an  alternative hypothesis when the results can be  attributed not to the reality.

So for this case a type I of error would be reject the hypothesis that the true mean is less or equal than 15 and is actually true.

3 0
2 years ago
Let C be the first quadrant portion of the circle of radius 3 centered on the origin. Find the line itegral for x^2+y^2 over C.
egoroff_w [7]
YX equals +2 squared minus -17 is approximately the answer of the origin
7 0
3 years ago
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