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Ivan
3 years ago
15

PLEASE HELP!!!‼️‼️‼️ I need help with question 14 a and b show your work please‼️

Mathematics
1 answer:
Anastaziya [24]3 years ago
6 0
A. The change after 1 hour is 2 inches because 6x10=60 (minutes) and you multiply 1/3 by 6/1 and get 6/3 or 2 inches.

b. The change after 2.25 hours is 4 1/2 inches because 2x2=4 and .25 (1/4) of an hour is 15 minutes so take 1/3 and multiply it by 1/2 (half of 10 is 5 and 10+5=15 min.) and you get 1/6 and then you add 1/3 and 1/6 together to get 1/2. So together it is 4 1/2 inches.
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432 students went on a field trip. Eight buses were filled with students. The
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Each bus was filled with 54 students
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5 0
3 years ago
Given the circle with equation x+ y =17
barxatty [35]

Answer:

Step-by-step explanation:

x+y=17 is NOT the equation for any circle. You have omitted the exponents. The equation for the circle should be

x^2 + y^2 = 17.

:::::

Plug the coordinates of D into the equation.

(-4)² + (-1)² = 17, so D is on the circle.

Plug the coordinates of E into the equation.

1² + 4² = 17, so E is on the circle.

:::::

Slope of DE = (-1 - 4)/(-4 - 1) = 1

Slope of perpendicular to DE is -1.

Midpoint of DE:  ((-4+1)/2, (-1+4)/2) = (-1.5, 1.5)

Point-slope equation for perpendicular bisector:

y-1.5 = -(x+1.5)

y = -x

:::::

Center of circle is (0,0)

0 = -0, so y=-x passes through the center.

8 0
3 years ago
Any athlete who fails the Enormous State University's women's soccer fitness test is automatically dropped from the team. Last y
castortr0y [4]

Using conditional probability, it is found that there is a 0.7873 = 78.73% probability that Mona was justifiably dropped.

Conditional Probability

P(B|A) = \frac{P(A \cap B)}{P(A)}

In which

  • P(B|A) is the probability of event B happening, given that A happened.
  • P(A \cap B) is the probability of both A and B happening.
  • P(A) is the probability of A happening.

In this problem:

  • Event A: Fail the test.
  • Event B: Unfit.

The probability of <u>failing the test</u> is composed by:

  • 46% of 37%(are fit).
  • 100% of 63%(not fit).

Hence:

P(A) = 0.46(0.37) + 0.63 = 0.8002

The probability of both failing the test and being unfit is:

P(A \cap B) = 0.63

Hence, the conditional probability is:

P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.63}{0.8002} = 0.7873

0.7873 = 78.73% probability that Mona was justifiably dropped.

A similar problem is given at brainly.com/question/14398287

3 0
2 years ago
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