What is k-15=-8?<br><br><br> I need help with this one<br> Btw it is asking what k equals.

-30 < x/-4 must be simplified

Norma-Jean [14]

Answer:

120 > x

Step-by-step explanation:

multiply -4 on both sides, -30 * -4 = 120

since you multiplied by a negative you change the sign

<3

7 0

3 years ago

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The time needed to complete a final examination in a particular college course is normally distributed with a mean of 77 minutes

Komok [63]

Answer:

a. The probability of completing the exam in one hour or less is 0.0783

b. The probability that student will complete the exam in more than 60 minutes but less than 75 minutes is 0.3555

c. The number of students will be unable to complete the exam in the allotted time is 8

Step-by-step explanation:

a. According to the given we have the following:

The time for completing the final exam in a particular college is distributed normally with mean (μ) is 77 minutes and standard deviation (σ) is 12 minutes

Hence, For X = 60, the Z- scores is obtained as follows:

Z= X−μ /σ

Z=60−77 /12

Z=−1.4167

Using the standard normal table, the probability P(Z≤−1.4167) is approximately 0.0783.

P(Z≤−1.4167)=0.0783

Therefore, The probability of completing the exam in one hour or less is 0.0783.

b. In this case For X = 75, the Z- scores is obtained as follows:

Z= X−μ /σ

Z=75−77 /12

Z=−0.1667

Using the standard normal table, the probability P(Z≤−0.1667) is approximately 0.4338.

Therefore, The probability that student will complete the exam in more than 60 minutes but less than 75 minutes is obtained as follows:

P(60<X<75)=P(Z≤−0.1667)−P(Z≤−1.4167)

=0.4338−0.0783

=0.3555

Therefore, The probability that student will complete the exam in more than 60 minutes but less than 75 minutes is 0.3555

c. In order to compute how many students you expect will be unable to complete the exam in the allotted time we have to first compute the Z−score of the critical value (X=90) as follows:

Z= X−μ /σ

Z=90−77 /12

Z=1.0833

UsING the standard normal table, the probability P(Z≤1.0833) is approximately 0.8599.

Therefore P(Z>1.0833)=1−P(Z≤1.0833)

=1−0.8599

=0.1401

Therefore, The number of students will be unable to complete the exam in the allotted time is= 60×0.1401=8.406

The number of students will be unable to complete the exam in the allotted time is 8

6 0

3 years ago

If f(x) = x-6 and g(x) = x^1/2 , find g(x) x f(x) .

shusha [124]

Answer:

x^3/2 - 6x^1/2

Step-by-step explanation:

f(x) = x - 6

g(x) = x^1/2

g(x) * f(x) = x^1/2(x - 6)

g(x) * f(x) = x^3/2 - 6x^1/2

5 0

3 years ago

How would I graph a line that contains the point (6,5) and has a slope of - 2/3 ?

ddd [48]

Answer:

y=-2/3x+9

Step-by-step explanation:

Use point-slope formula:

y-y1 = m(x-x1)

y-5=-2/3(x-6)

y-5=-2/3x+4

y=-2/3x+9

You would then proceed to graphing this equation, as it is in slope intercept form now.

HTH :)

6 0

3 years ago

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A certain virus infects one in every 300 people. A test used to detect the virus in a person is positive 90% of the time when t

garik1379 [7]

Answer:

a) P[A/B] = 0,019 or P[A/B] = 1,9 %

b) P[A- /B-] = 0,9996 or P[A- /B-] = 99,96 %

Step-by-step explanation:

Bayes Theorem :

P[A/B] = P(A) * P[B/A] / P(B)

The branches of events are as follows

Condition 1 real infection 1/300 and not infection 299/300

Then

1.- 1/300 299/300

When the test is done (virus present) 0,9 (+) 0,15 (-)

2.- 299/300

When the test is done ( no virus ) 0,15 (+) 0,85 (-)

Then:

P(A) = event person infected P(B) = person test positive

a) P[A/B] = P(A) * P[B/A] / P(B)

where P(A) = 1/300 = 0,0033 P[B/A] = 0,9

Then P(A) * P[B/A] = 0,0033*0,9 = 0,00297

P(B) is ( 1/300 )*0,9 + (299/300)*0,15

P(B) = 0,0033*0,9 + 0,9966*0,15 ⇒ P(B) = 0,1524

Finally

P[A/B] = 0,00297 /0,1524

P[A/B] = 0,019 or P[A/B] = 1,9 %

b) Following sames steps:

P[A- /B-] = (299/300) * 0,85 / (299/300) * 0,85 + (1/300 * 0,1)

P[A- /B-] = 0,8471 /0,8474

P[A- /B-] = 0,9996 or P[A- /B-] = 99,96 %

6 0

3 years ago

What is k-15=-8? I need help with this one Btw it is asking what k equals.