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MatroZZZ [7]
4 years ago
6

what are three cases where a limit does not exist? (& can you give ex of functions of graphs that represent these cases) ...

?
Mathematics
1 answer:
maks197457 [2]4 years ago
8 0
<span>1.) Jump discontinuities - Any line that jumps at a point
2.) Infinite discontinuities (y = 1/(x))
3.) Oscillating discontinuities (sin(x) as polpak suggested)</span>
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Sonia uses a gift card to purchase a $5.87 gourmet coffee each week. Find the total change in the gift card balance after 3 week
lianna [129]

Answer:

$17.61

Step-by-step explanation:

5.87 x 3= $17.61 because it is each week

6 0
3 years ago
Read 2 more answers
Un escalador sube 225m de altura. El asenso lo hizo en 3 etapas. En la primera subio un quinto en la, en la tercera un cuarto. ¿
kiruha [24]

Answer:

Segunda etapa= 123.75 metros

Step-by-step explanation:

Altura total= 225 metros

<u>En la primera estapa subió el 20% (un quinto):</u>

Primera etapa= 225*0.2= 45 metros

<u>En la tercera etapa subió 25% (un cuarto):</u>

Tercera etapa= 225*0.250 56.25 metros

Ahora debemos determinar cuánto subió en la segunda etapa:

Segunda etapa= altura total - total subido

Segunda etapa= 225 - (45 + 56.25)

Segunda etapa= 123.75 metros

6 0
3 years ago
What is the batting average of 34 out of 99
Gekata [30.6K]
<span>Safarina has 34 out of 99 hits.
Find her battling average.
=>  34 hits
=> 99 hits in all
x = value of percentage
=> 34 = x * 99
=> 34 = 99x
Now, let’s divide both sides with 99
=> 34 / 99 = 99x / 99
=> 0.3434343 = x (0.343434… is a repeating decimals, so need to round off to the nearest hundreds)
=> 0.34 = 34/100
Thus, the percentage of Sarafina’s hit is 34%</span>


7 0
3 years ago
Please help this is due today
Aliun [14]
I’m pretty sure it’s C.
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
4 years ago
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