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3 years ago

Whats 5000+35548-455x4534+38534x233=?

Mathematics

1 answer:

Vlada [557]3 years ago

6 0

Answer:

6,956,000

Do all the operations in order by following the PEMDAS method.

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When Anne had one cat, she needed to serve 1/8 of a can of cat

Alina [70]

Answer:

Abc= D= 2.94C

Step-by-step explanation:

4 0

3 years ago

At dominique's donuts the number of donut holes in a bag can vary. help dominique find the mode.12,10,10,10,13,12,11,13,10

sdas [7]

Hey!

To find the mode of a data set you'd have to observe the data set and then see which number occurs the most in the set. Now I could just let you see that on your own, but I'll provide a chart for you. I hope everything is 100% clear and properly labeled for you.

<em>PLEASE VIEW THE IMAGE PROVIDED</em>

Looking at the chart, you can see that the number 10 occurred 4 times and the other numbers occurred less than 2 times.

<em>The mode of the number of Dominique's donuts in her bag is</em> 10.

Hope this helps!

- Lindsey Frazier ♥

7 0

3 years ago

Consider a maximization linear programming problem with extreme points xi, x2, Xz. and x4. and extreme directions d1,. d2, and d

DIA [1.3K]

Answer:

Set of alternative optimal solution : 0 ≤ z ≤ 1.5

Hence There will be an infinite set of Alternative optimal solution

Step-by-step explanation:

considering Cx1 = 4

∴ C = 4 / x1

Cx2 = 6

∴ 4x2 - 6x1 = 0

2x2 - 3x1 = 0 ------ ( 1 )

considering Cx3 = 6

C = 6/x3

Cx4 = 3

∴ (6/x3) x4 - 3 = 0

= 2x4 - x3 = 0 ---- ( 2 )

attached below is the remaining part of the solution

set of alternative optimal solution : 0 ≤ z ≤ 1.5

There will be an infinite set of Alternative optimal solution

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3 years ago

What Could Medical Marijuana Help Cure?

Hunter-Best [27]

Cancer ,aids and more

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3 years ago

At a school with 100? students, 33 were taking? Arabic, 32 ?Bulgarian, and 40 Chinese. 9 students take only? Arabic, 12 take onl

Stels [109]

Answer: The answer is 4 and 32.

Step-by-step explanation: Let "A", "B" and "C" represents the set of students who were taking Arabic, Bulgarian and Chinese respectively.

The, according to the given information, we have

$n(A)=33,~~n(B)=32,~~n(C)=40,~~n(A\cap B)=14.$

Let 'p' represents the number of students who take all the three languages, then

$n(A\cap B\cap C)=p.$

Also,

$n(A)-n(A\cap B)-n(A\cap C)+n(A\cap B\cap C)=9\\\\\Rightarrow n(A\cap B)+n(A\cap C)=24+p~~~~~~~~~~~~~~(a),\\\\n(A\cap B)+n(B\cap C)=20+p~~~~~~~~~~~~~~(b),\\\\n(B\cap C)+n(A\cap C)=20+p~~~~~~~~~~~~~~~(c).$

From here, we get after subtracting equation(c) from (b) that

$n(A\cap B)=n(A\cap C)=14.$

Therefore,

$p=14+14-24=4,$ and from equation (a), we find

$n(B\cap C)=24-14=10.$

Thus,

$n(A\cap B\cap C)=4$ and

$n(A\cup B\cup C)=n(A)+n(B)+n(C)-n(A\cap B)-n(B\cap C)-n(A\cap C)+n(A\cap B\cap C)\\\\\Rightarrow n(A\cap B\cap C)=33+32+40-9-12-20+4\\\\\Rightarrow n(A\cap B\cap C)=68.$

Thus, the number of students who take all the three languages is 4 and the number of students who take none of the languages is 100-68 = 32.

5 0

3 years ago