All categories

4 years ago

The total of 26, 78, 198, 475 and 620 is

Mathematics

2 answers:

MissTica4 years ago

8 0

The answer is 1,397

26+78+198+475+620=1,397

zlopas [31]4 years ago

8 0

26+78+198+475+620=1397
The answer is 1397 .
Greetings!

You might be interested in

How many different committees can be formed from 12 teachers Abs 43 students of the committee consists of 4 teachers and 3 stude

Marina86 [1]

Answer:

6108795

Step-by-step explanation:

To obtain the number of ways the committee can be selected :

4 teachers from the 12 teachers available and 3 students from the 43 students available

Using combinatorics :

12C4 * 43C3

recall :

nCr = n! ÷ (n - r)!r!

Using calculator :

12C4 = 495

43C3 = 12341

12C4 * 43C3

495 * 12341

= 6108795

3 0

3 years ago

What is the domain..?please help!!

4vir4ik [10]

Answer:

The first choice.

Step-by-step explanation:

There are no real square roots of numbers < 0

so the domian is all real numbers greater than or equal to zero.

8 0

3 years ago

I need it badly as soon as possible if you can ASAP

Softa [21]

The answers are:

9:D

10:C

Have a good day! (or night, depending when u see this)

4 0

4 years ago

Enter the coefficient in the expression -2x+9

Monica [59]

Answer:

there is no

Step-by-step explanation:

4 0

3 years ago

Among 500 freshmen pursuing a business degree at a university, 323 are enrolled in an economics course, 205 are enrolled in a ma

Aleksandr-060686 [28]

<h3>The probability of selecting a random student who is enrolled in both the courses is 0.280.</h3>

Step-by-step explanation:

Here, the total number of freshman in the university = 500

The number of students enrolled in Economics = n(E) = 323

The number of students enrolled in Mathematics = n(M) = 205

The number of students enrolled in Both Economics and math

= n(E∩M ) = 140

Let F : Event of selecting a student who is enrolled in both the courses

So, from the given data:

$P(F) = \frac{\textrm{Total number of favorable outcomes}}{\textrm{Total number of outcomes}} \\= \frac{\textrm{Total number of students enrolled in BOTH courses}}{\textrm{Total number of students}} \\ = \frac{140}{500} = \frac{7}{25}$

So, the probability of selecting a random students who is enrolled in both the courses is $(\frac{7}{25} ) = 0.280$

4 0

3 years ago