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horsena [70]
2 years ago
12

Find the Value of X.

Mathematics
1 answer:
Paul [167]2 years ago
5 0

Answer:

  x = 50

Step-by-step explanation:

The marked angles are vertical angles, so are congruent.

  3x -3 = 147 . . . . congruent angles have equal measures

  x -1 = 49 . . . . . divide by3

  x = 50 . . . . . . add 1

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A jar is filled with only dimes and quarters. The jar contains a total of 60 coins and the value of the jar is $11.25. How many
inessss [21]

Answer:

35 quarters

Step-by-step explanation:

This situation has two unknowns - the total number of dimes and the total number of quarters. Because we have two unknowns, we will write a system of equations with two equations using the two unknowns.

  • d+q=60 is an equation representing the total number of coins
  • 0.10d+0.25q=11.25 is an equation representing the total value in money based on the number of coin. 0.10 and 0.25 come from the value of a dime and quarter individually.

We write the first equation in terms of q by subtracting it across the equal sign to get d=60-q. We now substitute this for d in the second equation.

0.10(60-q)+0.25q=11.25\\6-0.10q+0.25q=11.25\\6+0.15q=11.25

After simplifying, we subtract 6 across and divide by the coefficient of q.

6+0.15q=11.25\\0.15q=5.25\\q=35

We now know of the 60 coins that 35 are quarters. To find the total value of the quarters, we multiply 35 by 0.25 and find 8.75.

3 0
3 years ago
The math team does practice drills that each last 1/6 hour. In February the team did practice drills for a total of 24 hours. Ho
kozerog [31]

144 classes. each class is 10 minutes. 6X24= 144

7 0
4 years ago
A journal article reports that a sample of size 5 was used as a basis for calculating a 95% CI for the true average natural freq
oksano4ka [1.4K]

Answer:

Lower = 231.134- 3.098=228.036

Upper = 231.134+ 3.098=234.232

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

\bar X represent the sample mean for the sample  

\mu population mean (variable of interest)

s represent the sample standard deviation

n=5 represent the sample size  

Solution to the problem

The confidence interval for the mean is given by the following formula:

\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}   (1)

And for this case we know that the 95% confidence interval is given by:

\bar X=\frac{233.002 +229.266}{2}= 231.134

And the margin of error is given by:

ME = \frac{233.002 -229.266}{2}= 1.868

And the margin of error is given by:

ME= t_{\alpha/2} \frac{s}{\sqrt{n}}

The degrees of freedom are given by:

df = n-1 = 5-1=4

And the critical value for 95% of confidence is t_{\alpha/2}= 2.776

So then we can find the deviation like this:

s = \frac{ME \sqrt{n}}{t_{\alpha/2}}

s = \frac{1.868* \sqrt{5}}{2.776}= 1.506

And for the 99% confidence the critical value is: t_{\alpha/2}= 4.604

And the margin of error would be:

ME = 4.604 *\frac{1.506}{\sqrt{5}}= 3.098

And the interval is given by:

Lower = 231.134- 3.098=228.036

Upper = 231.134+ 3.098=234.232

6 0
3 years ago
5
Daniel [21]

Answer:

a,c,d

Step-by-step explanation:

8 0
2 years ago
What are the x and y-intercepts of the line described by the equation?
pishuonlain [190]
To find the x-intercept, substitute 0 for y and solve for x.
-6x + 3(0) = 18.9
-6x = 18.9
x = 18.9 / -6
x = -3.15 <-- Your x-intercept is (-3.15,0) 

To find the y-intercept, substitute in 0 for x and solve for y.
-6(0) + 3y = 18.9 
3y = 18.9
y = 18.9/3
y = 6.3 (0,6.3) <-- Your x-intercept is (0,6.3)

If this helped, please mark me brainliest, thank you and if you need more help, feel free to message me! 


8 0
3 years ago
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