Plz answer the question

The letters that form the word ALGEBRA are placed in a bowl. What is the probability, as a percent, of choosing a letter other t

KonstantinChe [14]

It will be 2/42 you count out the letters in “ALGEBRA” and then you can the A’s which equal a fraction 2/7 then you subtract 1 from the numerator and the denominator which then equals 1/6. Then you multiply the fractions and get 2/42

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

Brainliest for the correct answer!

liraira [26]

Answer:

y-3 = (1/4)x; 4y = -4x +14; y=4

Step-by-step explanation:

>First find the equation of function A

general equation of a line is y=mx+b

m = (y2-y1)/(x2-x1)= (0-(7/2))/(-7-0) = (-7/2) *(1/-7) = 1/2

consider the point (-7,0)

0 = -7(1/2) +b

7/2 = b

Function A has the equation y= (1/2)x +7/2

>rewrite the equations to fit the standard form

y-3 = (1/4)x ; add 3 to both sides

y= (1/4) x +3

4y = -4x +14 ,divide by 4 both sides

y= -x + 7/2

y=4

> intercept less than A

y-intercept for function A , y= (1/2)x +7/2, is 7/2 =3.5

y-intercept for function y= (1/4) x +3 is 3

3 < 7/2, so the answer is y-3 = (1/4)x

> intercept equal to A

y-intercept for function A , y= (1/2)x +7/2, is 7/2

y-intercept for function y= -x + 7/2 is 7/2

7/2=7/2 , so the answer is 4y = -4x +14

> intercept greater than A

y-intercept for function A , y= (1/2)x +7/2, is 7/2 =3.5

y-intercept for function y=4 is 4

4 > 7/2

6 0

3 years ago

Hello im back<br> i got a headache in Philippines server so i don't wanna come back to that server

raketka [301]

Answer:

Hello

Step-by-step explanation:

5 0

2 years ago

Read 2 more answers

If the demand function for a commodity is given by the equation

Jet001 [13]

Answer:

q=26.92 p=31.13

Step-by-step explanation:

$p^{2} +16q=1400\\\\q=\frac{1400-p^{2} }{16} ;\\\\700-p^{2} +10q=0\\\\q=\frac{p^{2}-700 }{10} ;\\\\$

$q=q\\\\\frac{1400-p^{2} }{16} =\frac{p^{2}-700 }{10} \\\\10(1400-p^{2})=16(p^{2}-700)\\\\14000-10p^{2}=16p^{2}-11200\\\\14000+11200=16p^{2}+10p^{2}\\\\25200=26p^{2}\\\\p^{2}=\frac{25200}{26} \\\\p=+\sqrt{\frac{25200}{26}} \\\\p=31.13$

replace p in any equation and

$q=\frac{1400-p^{2} }{16} \\\\q=\frac{1400-(31.13)^{2} }{16}\\\\q=26.92$

3 0

3 years ago