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finlep [7]
3 years ago
14

There are 4 apples and 5 oranges in a box. What is the probability of picking an apple and then an orange without replacement?

Mathematics
2 answers:
Annette [7]3 years ago
6 0
Pretty sure it's A.) 5/18 .
Fynjy0 [20]3 years ago
3 0
Probability of apple: 4/9
Probability of orange: 5/8
4/9*5/8=.2777777778
Which become 5/18 if you make it into a fraction. So A. :)
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30 pts + BRAINLIEST
ivanzaharov [21]

Answer:

Number of permutations = 5040

Step-by-step explanation:

We are given the word 'INNOVATIVE', that has total 10 letters.

Now, we are asked to form words using exactly 4 letters at a time i.e. out of 10 letters we are using 4.

Therefore, the number of permutations that can be formed using 4 letters are = P(10,4)

i.e. 10_{P_{4} }

= \frac{10!}{(10-4)!}

= \frac{10!}{6!}

= \frac{10*9*8*7*6!}{6!}

= 10*9*8*7

= 5040

8 0
3 years ago
Read 2 more answers
The difference of a number,n,and is less than 9
Soloha48 [4]

What do you need for this one???


6 0
3 years ago
A hiker sets out to climb a mountain with a 4,000 ft vertical ascent as shown in the diagram below. Ifthe angle of incline, a =
Ne4ueva [31]
hi fxgxgicgicigci log oh
6 0
2 years ago
HELP me plz someone!
Kamila [148]

Answer:

4,095 ft³

Step-by-step explanation:

V=LWH

V = 3 x 15 x 14

V= 4,095

4 0
2 years ago
Assume that you assign the following subjective probabilities for your final grade in your econometrics course (the standard GPA
rodikova [14]

Answer:

E(X) = 4*0.2 +3*0.5+ 2*0.2 +1*0.8+ 0*0.02= 2.78

So then the best answer for this case would be:

C. 2.78

Step-by-step explanation:

For this case we have the following probabability distribution function given:

Score         P(X)

A= 4.0       0.2

B= 3.0       0.5

C= 2.0       0.2

D= 1.0        0.08

F= 0.0       0.02

______________

Total          1.00

The expected value of a random variable X is the n-th moment about zero of a probability density function f(x) if X is continuous, or the weighted average for a discrete probability distribution, if X is discrete.

If we use the definition of expected value given by:

E(X) = \sum_{i=1}^n X_i P(X_I)

And if we replace the values that we have we got:

E(X) = 4*0.2 +3*0.5+ 2*0.2 +1*0.8+ 0*0.02= 2.78

So then the best answer for this case would be:

C. 2.78

3 0
3 years ago
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