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3 years ago

Find the value of x I’d appreciate if you could explain as well if not that’s fine

Mathematics

1 answer:

Kipish [7]3 years ago

7 0

All the angles in a circle equal to 360 degrees. So basically just add up all the known values: 23+23+15=61 and then subtract that from 360: 360-61=299.

You can also do 23+23+15+x=360. Adding it all becomes 61+x=360, subtract 61 from both sides of the equation, and you get x=299.

(Something to help remember is that the angle is obtuse, meaning it’s automatically more than 90 degrees, hope this helps!)

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If = and x > 0 what is the value of x ?

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3 years ago

The floor of hexagonal barn has a perimeter 1260

MArishka [77]

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3 years ago

Ahmed was invited at a party at his friend’s place at 20:00 hours. He left the house at 17:00 hours and travelled in his car at

WARRIOR [948]

Answer:

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3 years ago

How to work it out And who drove farther and how many miles farther

OlgaM077 [116]

Easy, just divide
kelly=440mi/8hr=220mi/4hr=110mi/2hr=55mi/1hr
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5 0

3 years ago

A researcher compares the effectiveness of two different instructional methods for teaching electronics. A sample of 138 student

yan [13]

Answer:

The 98% confidence interval for the true difference between testing averages for students using Method 1 and students using Method 2 is (-8.04, 0.84).

Step-by-step explanation:

The (1 - α)% confidence interval for the difference between population means is:

$CI=(\bar x_{1}-\bar x_{2})\pm z_{\alpha/2}\times \sqrt{\frac{\sigma^{2}_{1}}{n_{1}}+\frac{\sigma^{2}_{2}}{n_{2}}}$

The information provided is as follows:

$n_{1}= 138\\n_{2}=156\\\bar x_{1}=61\\\bar x_{2}=64.6\\\sigma_{1}=18.53\\\sigma_{2}=13.43$

The critical value of z for 98% confidence level is,

$z_{\alpha/2}=z_{0.02/2}=2.326$

Compute the 98% confidence interval for the true difference between testing averages for students using Method 1 and students using Method 2 as follows:

$CI=(\bar x_{1}-\bar x_{2})\pm z_{\alpha/2}\times \sqrt{\frac{\sigma^{2}_{1}}{n_{1}}+\frac{\sigma^{2}_{2}}{n_{2}}}$

$=(61-64.6)\pm 2.326\times\sqrt{\frac{(18.53)^{2}}{138}+\frac{(13.43)^{2}}{156}}\\\\=-3.6\pm 4.4404\\\\=(-8.0404, 0.8404)\\\\\approx (-8.04, 0.84)$

Thus, the 98% confidence interval for the true difference between testing averages for students using Method 1 and students using Method 2 is (-8.04, 0.84).

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3 years ago