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3 years ago

What is the LCD of 1/3 and 3/7?

Mathematics

1 answer:

Neko [114]3 years ago

3 0

The LCD of those fractions would be, 3 ... If this helped mark me brainliest. ty

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Stanley McBride is employed by the True Blue company. The PPO annual premium is $7,058. His employer pays 70% of the total cost.

Vitek1552 [10]

First deducte his contribution
7,058−7,058×0.70
=2,117.4
Then find weekly deduction
2,117.4÷52
=40.72

4 0

3 years ago

Read 2 more answers

What is the domain…..

saul85 [17]

Answer:

(-00, 00)

Step-by-step explanation:

(-infinity, infinity)

4 0

2 years ago

$10,000+8.55 sales tax

Pavlova-9 [17]

Answer:

$10,008.55

Add 10000, and 8.55 to give the total.

10,000.00

The cents will be .55 since the $10,000 does not have any cents included.

The $8 will stay since the $10,000 does not have any dollars below 10 dollars.

So, the answer will be $10,008.55

4 0

2 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

Tonya has 39 packages. Each package weighs 6 pounds. Choose the best estimate of the total weight of the packages.

Sidana [21]

Step-by-step explanation:

If were estimating, 39 is pretty close to 40 and 40x6 is 240 pounds. if we are getting exact about it, then 234 pounds

5 0

2 years ago