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Brrunno [24]
3 years ago
6

In an honors math class of 25 students, 15 are boys and 10 are girls. On the final exam, 6 boys and 3 girls received a grade of

A. If a student is chosen at
random from this class, what is the probability of choosing a girl or a student who did not receive an A on the exam?
1
0.76
2
0.64
3
0.40
0.28
Submit Answer
Mathematics
1 answer:
postnew [5]3 years ago
7 0

Answer:

Step-by-step explanation:

b=boy

a= a on the exam

b=15

b∩a=6

b'∩a=3

b'∪a'=?

b'∪a'= b'+a'-b'∩a'

a= b∩a+b'∩a (law of total probability) = 9

a'= 25-9 = 16

b'= 25-15= 10

16+10-7=19

19/25= .76

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A leprechaun places a magic penny under a​ girl's pillow. The next night there are 2 magic pennies under her pillow. The followi
vfiekz [6]

Answer:

34

Step-by-step explanation:

Since the pennies need to be kept under the pillow in order to be doubled, we're really looking for the day the girl will wake up with 5 billion pennies under her pillow.

The general formula of that geometric sequence will be: t = 1 + 2^(n-1)

We need to find the value of n when t will be over 5,000,000,000.

So, the equation becomes:

5,000,000,001 = 1 + 2^(n-1)

2^{n-1}= 5,000,000,001

log(2^{n-1}) = log(5,000,000,000)

(n-1) * log(2) = log(5,000,000,000)

n-1 = log(5,000,000,000) / log(2)

n-1 = 22.25

n = 33.25

Since it has to be complete days, and we have to reach over 5 billion, then we round it up to 34.

7 0
3 years ago
At what point is the function y=(x+3)/(x^(2)-7x+12) continuous
nika2105 [10]

If a function is defined as

h(x)=\dfrac{f(x)}{g(x)}

where both f(x),g(x) are continuous functions, then h(x) is also continuous where defined, i.e. where g(x)\neq0

So, in your case, this function is continous everywhere, except where

x^2-7x+12=0

To solve this equation, we can use the formula x^2-sx+p=0

It means that, if the leading terms is 1, then the x coefficient is the opposite of the sum of the roots, and the constant term is the product of the roots.

So, we're looking for two terms whose sum is 7, and whose product is 12. These numbers are easily found to be 3 and 4.

So, this function is continuous for every real number different than 3 or 4.

3 0
3 years ago
James is x years old now and the sum of his age is and his brothers age is 10.
Vesnalui [34]

Answer:

1) James brother is 10-x years old

2) James was 3 years ago x-3 years old

3) James brother was 3 years ago 7-x years old

4) 3(x-3)=7-x

5) James is 4 years old, his brother is 6 years old

Step-by-step explanation:

3) 10 - x - 3 = 7 - x

5) 3* James age 3 years ago = James brothers age 3 years ago

<=> 3*(x-3) = 7-x

<=> 3x - 9 = 7 - x

<=> 4x = 16

<=> x = 4

5 0
3 years ago
What is simplify 5+3-9
sergij07 [2.7K]

Answer:

-1

Step-by-step explanation:

first 5+3=8

8-9=-1

6 0
2 years ago
There is a triangle with a perimeter of 63 cm, one side of which is 21 cm. Also, one of the medians is perpendicular to one of t
NemiM [27]

Answer:

21cm; 28cm; 14cm

Step-by-step explanation:

There is no info in the problem/s  text which one of the triangle's  side is 21 cm. That is why we have to try all possible variants.

Let the triangle is ABC . Let the AK is the angle A bisector and BM is median.

Let O is AK and BM cross point.

Have a look to triangle ABM.  AO is the bisector and AOB=AOM=90 degrees (means that AO is as bisector as altitude)

=> triangle ABM is isosceles => AB=AM  (1)

1. Let AC=21   So AM=21/2=10.5 cm

So AB=10.5 cm as well.  So BC= P-AB-AC=63-21-10.5=31.5 cm

Such triangle doesn' t exist ( is impossible), because the triangle's inequality doesn't fulfill.  AB+AC>BC ( We have 21+10.5=31.5 => AB+AC=BC)

2. Let AB=21 So AM=21 and AC=42 .So  BC= P-AB-AC=63-21-42=0 cm- such triangle doesn't exist.

3. Finally let BC=21 cm. So AB+AC= 63-21=42 cm

We know (1) that AB=AM so AC=2*AB.  So AB+AC=AB+2*AB=3*AB

=>3*AB=42=> AB=14 cm => AC=2*14=28 cm.

Let check if this triangle exists ( if the triangle's inequality fulfills).

BC+AB>AC    21+14>28 - correct=> the triangle with the sides' length 21cm,14 cm, 28cm exists.

This variant is the only possible solution of the given problem.

6 0
3 years ago
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