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2 years ago

A student says that the sum difference, product and quotient of two

Mathematics

1 answer:

anastassius [24]2 years ago

7 0

Answer:

The quotient of two integers may not always be an integer.

Therefore, I do not agree when a student says that the sum difference, product, and quotient of two are always integers.

Step-by-step explanation:

The student is not largely correct!

The sum, difference, and product of two integers is indeed always an integer.

But, the quotient of two integers may not always be an integer.

For example, the quotient of integers 4 and 2 will be an integer.

i.e.

4/2 = 2

But, if we take the quotient of 2 and 3, the result will not be an integer.

i.e.

2/3 = 0.67

Therefore, I do not agree when a student says that the sum difference, product, and quotient of two are always integers.

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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.

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On a cold February morning, the radiator fluid in Stanley’s car is -18 degrees When the engine is running, the temperature goes

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Remark

This is an arithmetic progression. You have the first and last terms and the difference.

Givens

a = - 18

L = 60

d = 4.5

Equation

L = a + (n - 1)d

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60 = -18 + (n - 1) * 4.5 Add 18 to both side

60 + 18 = (n - 1) * 4.5

78 = (n - 1) * 4.5 Divide both sides by 4.5

78/4.5 = n - 1

17.3333 = n - 1 Add 1 to both sides.

18.3333 = n

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3 years ago

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victus00 [196]

Answer:

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(x - 3)²

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3 years ago

if a sequence is defined recursively by f(0)=4 and f(n+1)= -3f(n+1)=-3f(n)+1 for n≥0, then f(3) is equal to? DESPERATELY NEED HE

muminat

f(3) is equal to -101.

<u>Step-by-step explanation</u>:

The sequence is followed by f(0)=4
The given expression is f(n+1)= -3f(n)+1

Put n=0,

f(0+1)= -3f(0)+1

f(1)= -3(4)+1

f(1)= -12+1 = -11

Put n=1,

f(1+1)= -3f(1)+1

f(2)= -3(-11)+1

f(2)= 33+1 = 34

Put n=2,

f(2+1)= -3f(2)+1

f(3)= -3(34)+1

f(3)= -102+1

f(3)= -101

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3 years ago