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3 years ago

Find all possible values of k so that x^2 + kx - 32 can be factored.

Mathematics

1 answer:

daser333 [38]3 years ago

8 0

Answer:

Step-by-step explanation:

The posssible factors are

x^2 + 4x - 32

(x+8)(x-4)

x^2+ 14x - 32

(x+16)(x-2)

(x+32)(x-1)

x^2+31x-32

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A new screening test for Lyme disease is developed for use in the general population. The sensitivity and specificity of the new

Anit [1.1K]

Answer:

predictive value of a positive test = 18.18%

predictive value of a negative test = 94.03%

Step-by-step explanation:

Sensitivity = 60% = 0.6

Specificity = 70% = 0.7

Let True Positive = TP

True Negative = TN

False Negative = FN

$Sensitivity = \frac{TP}{TP + FN} \\0.6 = \frac{TP}{TP + FN} \\0.6TP + 0.6FN = TP\\0.4TP = 0.6FN\\TP = 1.5 FN$

$Specificity = \frac{TN}{TN + FP} \\0.7 = \frac{TN}{TN + FP} \\0.7TN + 0.7FP = TN\\0.7FP = 0.3 TN\\TN = 7/3 FP$

Prevalence = 10% = 0.1

Three hundred people are screened, $T_{total} = 300$

Total number of people having the disease, $T_{disease} = ?$

$Prevalence = \frac{T_{disease} }{T_{total} } \\0.1 = \frac{T_{disease} }{300 }\\T_{disease} = 30$

$T_{disease} = TP + FN\\30 = TP + FN$

But TP = 1.5 FN

30 = 1.5 FN + FN

30 = 2.5 FN

FN = 30/2.5

FN = 12

TP = 1.5 FN = 1.5 * 12

TP = 18

$FP + TN = T_{total} - T_{disease} \\FP + TN = 300 - 30\\FP + TN = 270\\FP + \frac{7}{3} FP = 270\\\frac{10}{3} FP = 270\\FP = 27 * 3\\FP = 81$

81 + TN = 270

TN = 189

To calculate the Predictive value of positive test (PPT)

$PPT = \frac{TP}{TP + FP} * 100\\PPT = \frac{18}{18+81} * 100\\PPT = \frac{18}{99} * 100\\PPT = 18.18 \%$

To calculate the Predictive value of negative test (PNT)

$PPT = \frac{TN}{FN + TN} * 100\\PPT = \frac{189}{189+12} * 100\\PPT = \frac{189}{201} * 100\\PPT = 94.03 \%$

7 0

2 years ago

2x^2+3x-2, a=0 How do I find the derivative?

Ghella [55]

You can use the definition:

$\displaystyle f'(x) = \lim_{h\to0}\frac{f(x+h)-f(x)}h$

Then if

$f(x) = 2x^2+3x-2$

we have

$f(x+h) = 2(x+h)^2+3(x+h) - 2 = 2x^2 + 4xh+2h^2+3x+3h-2$

Then the derivative is

$\displaystyle f'(x) = \lim_{h\to0}\frac{(2x^2+4xh+2h^2+3x+3h-2)-(2x^2+3x-2)}h \\\\ f'(x) = \lim_{h\to0}\frac{4xh+2h^2+3h}h \\\\ f'(x) = \lim_{h\to0}(4x+2h+3) = \boxed{4x+3}$

I'm guessing the second part of the question asks you to find the tangent line to f(x) at the point a = 0. The slope of the tangent line to this point is

$f'(0) = 4(0) + 3 = 3$

and when a = 0, we have f(a) = f (0) = -2, so the graph of f(x) passes through the point (0, -2).

Use the point-slope formula to get the equation of the tangent line:

y - (-2) = 3 (x - 0)

y + 2 = 3x

y = 3x - 2

5 0

2 years ago

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SVETLANKA909090 [29]

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2 years ago

Show the tens fact you used. Write the difference. 16-9= 10-___=_____

gayaneshka [121]

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2 years ago

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jeka57 [31]

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2 years ago