Find the value of x and y. Note: Input answer as x, y

Write the point-slope equation of the line that goes through (1, 4) with a slope of -10

sammy [17]

Answer:

y-4=-10(x-1)

Step-by-step explanation:

Point-slope form is:

$y-y_1=m(x-x_1)\\m-slope\\(x_1,y_1)-point$

We are given the slope of -10 and the point (1,4). So:

m = -10

x1 = 1

y1= 4

Replace the appropriate values:

$y-4=-10(x-1)$

So, y-4=-10(x-1) should be the point-slope equation.

3 0

2 years ago

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A shipment of 50 precision parts including 4 that are defective is sent to an assembly plant. The quality control division selec

natali 33 [55]

Answer:

0.3968 = 39.68% probability this shipment passes inspection.

Step-by-step explanation:

The parts are chosen without replacement, which means that the hypergeometric distribution is used to solve this question.

Hypergeometric distribution:

The probability of x successes is given by the following formula:

$P(X = x) = h(x,N,n,k) = \frac{C_{k,x}*C_{N-k,n-x}}{C_{N,n}}$

In which:

x is the number of successes.

N is the size of the population.

n is the size of the sample.

k is the total number of desired outcomes.

Combinations formula:

$C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

In this question:

50 parts means that $N = 50$

4 defective means that $k = 4$

10 are chosen, which means that $n = 10$

What is the probability this shipment passes inspection?

Probability that none is defective, so:

$P(X = x) = h(x,N,n,k) = \frac{C_{k,x}*C_{N-k,n-x}}{C_{N,n}}$

$P(X = 0) = h(0,50,10,4) = \frac{C_{4,0}*C_{46,10}}{C_{50,10}} = 0.3968$

0.3968 = 39.68% probability this shipment passes inspection.

6 0

2 years ago

Select three expressions equivalent to -3/4

Daniel [21]

Answer:

0.75 as a decimal for -3/4

I hope this helps

5 0

3 years ago

Three times a number plus five equals twenty

gladu [14]

3x + 5 = 20

x = 5

Hope this helps

8 0

3 years ago

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PLEASE PLEASE HELP! Find the measure of APB^.

Whitepunk [10]

Answer:

65°

Step-by-step explanation:

Radii CA and CB are perpendicular to tangent lines AT and BT, so

$\angle CAT=\angle CBT=90^{\circ}$

Since angle BAT is equal to 65°, angle CAB has measure

$\angle CAB=90^{\circ}-65^{\circ}=25^{\circ}.$

Consider triangle ACB. This triangle is isosceles, because CA=CB as radii of the circle. Two angles adjacent to the base are congruent, thus

$\angle CBA=\angle CAB=25^{\circ}$

The sum of the measures of all interior angles in triangle is always 180°, so

$\angle CAB+\angle CBA+\angle ACB=180^{\circ}\\ \\25^{\circ}+25^{\circ}+\angle ACB=180^{\circ}\\ \\\angle ACB=130^{\circ}$

Angle ACB is central angle subtended on the minor arc AB, angle APB is inscribed angle subtended on the same minor arc AB. The measure of inscribed angle is half the measure of central angle subtended on the same arc, so

$x=\angle APB=\dfrac{1}{2}\cdot 130^{\circ}=65^{\circ}$

8 0

3 years ago