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2 years ago

Is 24/40 4/5 a true proportion?

Mathematics

1 answer:

cupoosta [38]2 years ago

5 0

Answer:

Step-by-step explanation:

24 x 5 = 120

40 x 4= 160

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A large bag of marbles contains red (R) and blue (B) marbles. Marbles are pulled one at a time, recorded, and placed back in the

ehidna [41]

The answer will be A.

6 0

1 year ago

A dartboard has 10 equally sized slices numbered from 1 to 10. Some are grey and some are white. The slices numbered 1,2 ,3 ,5 ,

hram777 [196]

Answer:

P(X) is 7/10

P(not X) is 3/10

Step-by-step explanation:

The total is 10 so thats ur denominator

The probability of landing on a grey area is 7/10

The probability of not landing on a grey area is 3/10

Hope this helps

5 0

1 year ago

Find any relative extrema of the function. (Round your answers to three decimal places.)

Strike441 [17]

(x) = arcsec(x) − 8x

f'(x) = d/dx( arcsec(x) − 8x )

1/xsqrt( x^2 - 1) - 8

f'(x) = 0

1/xsqrt( x^2 - 1) - 8 = 0

8 x sqrt (x^2-1) = 1

( 8 x sqrt (x^2-1) )^2 = 1

64 x^2 ( x^2 - 1) = 1

64 x^4 - 64 x^2 =1

64 x^4 - 64 x^2 - 1 = 0

x = 1.00766 , - 1.00766

x = - 1.00766

f(- 1.00766) = arcsec(- 1.00766) − 8( - 1.00766)

f( - 1.00766 ) = 11.07949

x = 1.00766

f(1.00766) = arcsec(1.00766) − 8( 1.00766)

f(1.00766 ) = -7.93790

relative maximum (x, y) = (- 1.00766 , 11.07949 ) relative minimum (x, y) = ( 1.00766 , -7.93790 )

8 0

2 years ago

Consider the midterm and final for a statistics class. Suppose 13% of students earned an A on the midterm. Of those students who

padilas [110]

Answer:

There is a 38.97% probability that this student earned an A on the midterm.

Step-by-step explanation:

The first step is that we have to find the percentage of students who got an A on the final exam.

Suppose 13% students earned an A on the midterm. Of those students who earned an A on the midterm, 47% received an A on the final, and 11% of the students who earned lower than an A on the midterm received an A on the final.

This means that

Of the 13% of students who earned an A on the midterm, 47% received an A on the final. Also, of the 87% who did not earn an A on the midterm, 11% received an A on the final.

So, the percentage of students who got an A on the final exam is

$P_{A} = 0.13(0.47) + 0.87(0.11) = 0.1568$

To find the probability that this student earned an A on the final test also earned on the midterm, we divide the percentage of students who got an A on both tests by the percentage of students who got an A on the final test.

The percentage of students who got an A on both tests is:

$P_{AA} = 0.13(0.47) = 0.0611$