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Morgarella [4.7K]
3 years ago
13

Using the numbers 8,6,4 and 2 write an expression that equals 4

Mathematics
2 answers:
Zolol [24]3 years ago
6 0
6+4-8+2 = 4 is the answer
vfiekz [6]3 years ago
5 0
8-(6+2)+4 is how you use those numbers to equal 4
You might be interested in
Write the linear equation in slope-intercept form passing through the point (0, 3) and that has a slope of 2.
Yuri [45]

Answer:

y=2x+3

Step-by-step explanation:

5 0
3 years ago
Miriam is picking out some movies to rent, and she is primarily interested in comedies and foreign films. She has narrowed down
Keith_Richards [23]

Answer:

There are 795 combinations.

Step-by-step explanation:

The number of ways or combinations in which we can select k element from a group of n elements is given by:

nCk=\frac{n!}{k!(n-k)!}

So, if Miriam want to choose 3 movies with at least two comedies, she have two options: Choose 2 comedies and 1 foreign film or choose 3 comedies.

Then, the number of combinations for every case are:

1. Choose 2 Comedies from the 10 and choose 1 foreign film from 15. This is calculated as:

10C2*15C1=\frac{10!}{2!(10-8)!}*\frac{15}{1!(15-14)!}

10C2*15C1=675

2. Choose 3 Comedies from the 10. This is calculated as:

10C3=\frac{10!}{3!(10-3)!}=120

Therefore, there are 795 combinations and it is calculated as:

675 + 120 = 795

8 0
3 years ago
Please help ASAP!
Lana71 [14]
1)x+1) has positive slope, it is only A or B
2) For f(x)=4 we have x≥1, so this graph should begin with black dot.
3)So answer is B
Blue dot should be empty.
Orange dot should be black.

8 0
3 years ago
URGENT!!!!
Simora [160]

Answer:

cross multipy them into a fraction

Step-by-step explanation:


5 0
3 years ago
Ignoring leap days, the days of the year can be numbered 1 to 365. Assume that birthdays are equally likely to fall on any day o
faust18 [17]

Answer:

Follows are the solution to the given question:

Step-by-step explanation:

They can count the days of the year 1 to 365. The random project consists of drawing a sample of n objects from D where elements are n people's birth in a group but instead, D = {1,....365}. And then there's the issue.

S=365^n

This because the list of future birthdays of n people was its test point; therefore m points will be in the sequence so each point contains 365 distinct outcomes. The probability function P for \Omega is that any event is likely to happen in 365 days.

P(x)=\frac{1}{365^{n}}

if x is between 1 and 365 as well as the occurrence is just all similarly possible

In point i:

That somebody mentions their birthday throughout the party

Guess I was born on day b. Therefore the consequence of "x is in A" is "b is now in the series of x," which is to say, b = bk for some amount k approximately 1 and n.

In point ii:

Any 2 persons share the same birthday at this party". A result x is in B" means which "two of entries in x are same." This means that perhaps the outcome x is in B if or only if bj = bk is in B of two numbers j, and k of 1, of two. , no, n.

In point iii:

Many three students share the same birthday with both the party. The consequence is x at the level of C only when bj = bk = bl at three (different) indices, j, k, l, 1. , no, n.

6 0
3 years ago
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