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3 years ago

Someone please help me with this√2w-t / 3 = 6 [w]

Mathematics

1 answer:

Sergio [31]3 years ago

7 0

Answer:w=6+t/3/2 squared

Step-by-step explanation:

2 squared w= 6+t/3

w=6+t/3/2 squared

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X +3y =15 x-2y=10 help plz

solong [7]

Answer:

What is that

Step-by-step explanation:

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3 years ago

Read 2 more answers

BRAINLIEST +10 POINTS

Oksanka [162]

The forth one I can’t type that tho

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3 years ago

The value of X is ...

Alex787 [66]

Step-by-step explanation:

sums of external angles of triangle =360°

360 - 130 -134

x = 96°

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3 years ago

Does 24, 7, and 26 form a right triangle

const2013 [10]

Answer:

No. It is an obtuse triangle.

Step-by-step explanation:

If c^2>a^2+b^2, then the triangle is obtuse. Since 24^2 + 7^2 is 625, and the square root of 625 is 25, that means 26>25. This proves that the triangle is obtuse.

4 0

3 years ago

Suppose twenty-two communities have an average of = 123.6 reported cases of larceny per year. assume that σ is known to be 36.8

Delvig [45]

We are given the following data:

Average = m = 123.6
Population standard deviation = σ= psd = 36.8
Sample Size = n = 22

We are to find the confidence intervals for 90%, 95% and 98% confidence level.

Since the population standard deviation is known, and sample size is not too small, we can use standard normal distribution to find the confidence intervals.

Part 1) 90% Confidence Interval
z value for 90% confidence interval = 1.645

Lower end of confidence interval = $m-z *\frac{psd}{ \sqrt{n} }$
Using the values, we get:
Lower end of confidence interval= $123.6-1.645* \frac{36.8}{ \sqrt{22}}=110.69$

Upper end of confidence interval = $m+z *\frac{psd}{ \sqrt{n} }$
Using the values, we get:
Upper end of confidence interval= $123.6+1.645* \frac{36.8}{ \sqrt{22}}=136.51$

Thus the 90% confidence interval will be (110.69, 136.51)

Part 2) 95% Confidence Interval
z value for 95% confidence interval = 1.96

Lower end of confidence interval = $m-z *\frac{psd}{ \sqrt{n} }$
Using the values, we get:
Lower end of confidence interval= $123.6-1.96* \frac{36.8}{ \sqrt{22}}=108.22$

Upper end of confidence interval = $m+z *\frac{psd}{ \sqrt{n} }$
Using the values, we get:
Upper end of confidence interval= $123.6+1.96* \frac{36.8}{ \sqrt{22}}=138.98$

Thus the 95% confidence interval will be (108.22, 138.98)

Part 3) 98% Confidence Interval
z value for 98% confidence interval = 2.327

Lower end of confidence interval = $m-z *\frac{psd}{ \sqrt{n} }$
Using the values, we get:
Lower end of confidence interval= $123.6-2.327* \frac{36.8}{ \sqrt{22}}=105.34$
Upper end of confidence interval = $m+z *\frac{psd}{ \sqrt{n} }$
Using the values, we get:
Upper end of confidence interval= $123.6+2.327* \frac{36.8}{ \sqrt{22}}=141.86$

Thus the 98% confidence interval will be (105.34, 141.86)

Part 4) Comparison of Confidence Intervals
The 90% confidence interval is: (110.69, 136.51)
The 95% confidence interval is: (108.22, 138.98)
The 98% confidence interval is: (105.34, 141.86)

As the level of confidence is increasing, the width of confidence interval is also increasing. So we can conclude that increasing the confidence level increases the width of confidence intervals.

3 0

3 years ago

Someone please help me with this√2w-t / 3 = 6 [w]​

Someone please help me with this√2w-t / 3 = 6 [w]