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2 years ago

The product of three consecutive integers is 210. What is their sum?

Mathematics

1 answer:

Tcecarenko [31]2 years ago

6 0

Answer:

3 consecutive no.s are

5×6×7=210

sum of them is= 5+6+7

=18

So, I hope. it helps !!

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Find the missing length <br> . Leave your answers in simplest radical form

Aloiza [94]

A^2 + B^2 = C^2

x^2 + 9^2 = (SQRT 226)^2

x^2 + 81 = 226

x^2 = 145

x = SQRT 145

8 0

3 years ago

Fractor 48-8x to identify the equivalent expressions

ale4655 [162]

Answer:

8(6-x)

Step-by-step explanation:

factor out an 8

8*6 -8*x

8(6-x)

8 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

What is the product of the polynomials below? (6x3 + 3x)(x2 + 4)

shutvik [7]

6x5+24x3+3x3+12x ------> Simplify by adding like terms

6x5 + 27x3 + 12x

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3 years ago

Translate this sentence into an equation.<br><br> The difference of Mai's age and 12 is 60

daser333 [38]

Answer: 72

Step-by-step explanation:

7 0

3 years ago

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