The graph of which equation is shown below?

Jan had $126. she spent 1/6 of her money on her makeup and 2/3 of her money on shoes. How Much does Jan have left?

Jobisdone [24]

She has 1/6 of her money left, which is $21.

2/3=4/6, so 4/6 + 1/6=5/6, which means she has 1/6 left. Divide $126 by 6 and you get $21.

3 0

3 years ago

Read 2 more answers

PLEASE HELP I WILL MARK BRAINLIEST

Mice21 [21]

Answer:

180 i think if you are not sure with my ans then just don't use it k bey

Step-by-step explanation:

7 0

3 years ago

Solve for x in the equation 9- x/4=2

jeka57 [31]

Answer:

9-x/4=2

$\frac{36 - x}{4} = 2$

$36 - x = 2 \times 4$

$36 - x = 8$

$36 - 8 = x$

$28 = x$

$x = 28$

5 0

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

show that the curve has 3 points of inflection and they all lie on 1 straight line:\[y=\frac{1+x}{1+x^{2}}\]

ohaa [14]

Y = (1 + x) / (1 + x^2)

y'
= [(1 + x^2)(1) - (1 + x)(2x)] / (1 + x^2)^2
= [1 + x^2 - 2x - 2x^2] / (1 + x^2)^2
= [-x^2 - 2x + 1] / (1 + x^2)^2

y''
= [(1 + x^2)^2 * (-2x - 2) - (-x^2 - 2x + 1)(2)(1 + x^2)(2x)] / (1 + x^2)^4
= [(1 + x^2)(-2x - 2) - (4x)(-x^2 - 2x + 1)] / (1 + x^2)^3
= [(-2x - 2x^3 - 2 - 2x^2) - (-4x^3 - 8x^2 + 4x)] / (1 + x^2)^3
= [-2x - 2x^3 - 2 - 2x^2 + 4x^3 + 8x^2 - 4x] / (1 + x^2)^3
= [2x^3 + 6x^2 - 6x - 2] / (1 + x^2)^3

Setting y'' to zero, we have:
y'' = 0
[2x^3 + 6x^2 - 6x - 2] / (1 + x^2)^3 = 0
(2x^3 + 6x^2 - 6x - 2) = 0

Using trial and error, you will realise that x = 1 is a root.
This means (x - 1) is a factor.
Dividing 2x^3 + 6x^2 - 6x - 2 by x - 1 using long division, you will have 2x^2 + 8x + 2.

2x^2 + 8x + 2
= 2(x^2 + 4x) + 2
= 2(x + 2)^2 - 2(2^2) + 2
= 2(x + 2)^2 - 8 + 2
= 2(x + 2)^2 - 6

Setting 2x^2 + 8x + 2 to zero, we have:
2(x + 2)^2 - 6 = 0
2(x + 2)^2 = 6
(x + 2)^2 = 3
x + 2 = sqrt(3) or = -sqrt(3)
x = -2 + sqrt(3) or x = -2 - sqrt(3)

Note that -2 - sqrt(3) < -2 + sqrt(3) < 1
We will choose random values belonging to each interval and test them out.

-5 < -2 - sqrt(3) < -2 < -2 + sqrt(3)
f''(-5) = [2(-5)^3 + 6(-5)^2 - 6(-5) - 2] / (1 + (-5)^2)^3 = -9/2197 < 0
f''(-2) = [2(-2)^3 + 6(-2)^2 - 6(-2) - 2] / (1 + (-2)^2)^3 = 18/125 > 0
Note that one value is positive and the other is negative.
Thus, x = -2 - sqrt(3) is an inflection point.

-2 - sqrt(3) < -2 < -2 + sqrt(3) < 0 < 1
f''(-2) = [2(-2)^3 + 6(-2)^2 - 6(-2) - 2] / (1 + (-2)^2)^3 = 18/125 > 0
f''(0) = [2(0)^3 + 6(0)^2 - 6(0) - 2] / (1 + (0)^2)^3 = -2 < 0
Note that one value is positive and the other is negative.
Thus, x = -2 + sqrt(3) is also an inflection point.

-2 + sqrt(3) < 0 < 1 < 2
f''(0) = [2(0)^3 + 6(0)^2 - 6(0) - 2] / (1 + (0)^2)^3 = -2 < 0
f''(2) = [2(2)^3 + 6(2)^2 - 6(2) - 2] / (1 + (2)^2)^3 = 26/125 > 0
Note that one value is positive and the other is negative.
Thus, x = 1 is an inflection point.

Hence, we have three inflection points in total.

When x = -2 - sqrt(3), we have:
y
= (1 - 2 - sqrt(3)) / (1 + (-2 - sqrt(3))^2)
= (-1 - sqrt(3)) / (1 + 4 + 4sqrt(3) + 3)
= (-1 - sqrt(3)) / (8 + 4sqrt(3))

When x = -2 + sqrt(3), we have:
y
= (1 - 2 + sqrt(3)) / (1 + (-2 + sqrt(3))^2)
= (-1 + sqrt(3)) / (1 + 4 - 4sqrt(3) + 3)
= (-1 + sqrt(3)) / (8 - 4sqrt(3))

When x = 1, we have:
y
= (1 + 1) / (1 + 1^2)
= 2 / 2
= 1

Using the slope formula, we have:
(y - 1) / (x - 1) = [[(-1 + sqrt(3)) / (8 - 4sqrt(3))] - 1] / ( -2 + sqrt(3) - 1)
(y - 1) / (x - 1) = 1/4, which is the equation of the line which the inflection points at x = 1 and x = -2 + sqrt(3) lies on.

Note that I am skipping the intermediate steps for simplifying here, but the trick is to rationalise the denominator by multiplying a conjugate on both numerator and denominator.

Now, we just need to check that the inflection point at x = -2 - sqrt(3) lies on the same line as well.
L.H.S.
= [[(-1 - sqrt(3)) / (8 + 4sqrt(3))] - 1] / (-2 - sqrt(3) - 1)
= 1/4
= R.H.S.

Once again, I am skipping simplifying steps here.

<span>Anyway, this proves all three points of inflection lies on the same straight line.</span>

4 0

3 years ago