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3 years ago

Plz helppppp i am struggling

Mathematics

1 answer:

dimulka [17.4K]3 years ago

4 0

Answer:

$\left(\frac{h}{g}\right)(a) = \frac{a^2-2}{3a-1}$ for $a \neq \frac{1}{3}$ .

Step-by-step explanation:

By definition, $\left(\frac{h}{g}\right)(a)$ is equal to $\frac{h(a)}{g(a)}$ , which is equal to $\frac{a^2-2}{3a-1}$ where $a \neq \frac{1}{3}$ . If $a = \frac{1}{3}$ , then $g(a)$ is equal to zero and $\left(\frac{h}{g}\right)(a)$ is not defined.

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PLEASE HELP, Find the slope of the following lines :D

joja [24]

Number one is 2/6 number 2 is -10/4 and number three is 4.
you have to do rise over run and then divide if you need to. if it’s going to the right then it’s positive and if it’s going to the left then it’s negative.

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4 years ago

Dan invests £1200 into his bank account. He receives 5% per year compound interest. How much will dan have after5 years?

solmaris [256]

Answer:

Dan will have $1,531.53 after 5 years.

Step-by-step explanation:

To find the answer, you can use the following formula to calculate the future value:

F= P(1 + r)^t

F= Future value

P= Present value= 1200

r= rate of interest= 5%

t= time= 5

F=1200(1+0.05)^5

A=1200(1.05)^5

A=1531.53

According to this, the answer is that Dan will have $1,531.53 after 5 years.

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3 years ago

When dealing with word problems and linear equations, what does the y-intercept represent and how is the slope defined?

katovenus [111]

Answer:

Step-by-step explanation:

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6 0

3 years ago

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In a week karry ran 9.3 miles and Tricia ran 4.4 miles.Use mental math to find how much farther Karry ran than Tricia.Explain ho

sineoko [7]

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3 years ago

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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

4 years ago