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4 0

He has 2.5 days worth of seed left. 3/3 or 6/6 equals three days worth of seed, and 2/6 equals one day worth of seed, so one-half a day (1/6) less is 2.5 days, or 5/6.

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Change 108% to a mixed number and reduce, if possible.

monitta

27/25 is in the lowest form

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Colby deposited money into a savings account that pays a simple annual interest of 1.8%. He earned $334.80 in interest after 6 y

kicyunya [14]

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Step-by-step explanation:

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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

Linear equations 3(x+1)=3x+3

vaieri [72.5K]

4x + 2 - 3x + 5 - 2(x + 5) = (2x - 5) + (3x + 4)
4x + 2 - 3x + 5 - 2x - 10 = 2x - 5 + 3x + 4
4x - 3x - 2x + 5 + 2 - 10 = 2x + 3x - 5 + 4
- x - 3 = 5x - 1
-x - 3 = 5x - 1 is the line above recopied
+x = +x
   -3 = 6x - 1
<span>   +1 =    + 1 </span>
   - 2 = 6x
   - 2/6 = 6x/6
   - 1/3 = x, the answer

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3 years ago

Dy/dx=(x+1)^2 in problem solve the given differential equation by separation of variable

Brilliant_brown [7]

The derivative of (x+1)^2, for this you use the chain rule, to get 2x+2.

3 0

3 years ago