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3 years ago

Find the sum of 268 and 479

Mathematics

2 answers:

Readme [11.4K]3 years ago

8 0

Answer:

747

Step-by-step explanation:

Tina has 268 brainly points, she answers a question with 479 brainly points. How many points does she have now?

Answer - 747

Hope this helps fam!

~CoCo <3

mamaluj [8]3 years ago

5 0

Answer:

The sum of 268 and 479 is 747.

Step-by-step explanation:

Find the sum of 268 and 479

To find the sum means we have to add the two numbers.

So, we will add $268 and 479$

$268+479$ = 747

Hence, The sum of 268 and 479 is 747.

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10 coupons are given to 20 shoppers in a store. Each shopper can receive at most one coupon. 5 of the shoppers are women and 15

zmey [24]

Answer:

$\, {20 \choose 10} - {15 \choose 10}$

Step-by-step explanation:

We can find the total amount of ways at least a woman receives a coupon by calculating the total amount of possibilities ot distribute the coupon and substract it to the total amount of possibilities to distribute 10 coupons to the 15 men (this is the complementary case that at least a woman receives a cupon).

- The total amount of possibilities to distribute the coupons among the 20 shoppers is equivalent to the total amount of ways to pick a subset of 10 elements from a set of 20. This is the combinatiorial number of 20 with 10, in other words, $\, {20 \choose 10} = \frac{20!}{10!10!} .$

- To calculate the total amount of possibilities to distribute the coupons among the 15 men, we need to make the same computation we made above but with a set of 15 elements instead of 20. This gives us $\, {15 \choose 10} = \frac{15!}{10!5!}$ possibilities.

Therefore, we have $\, {20 \choose 10} - {15 \choose 10}$ possibilities to distribute the coupons so that at least one woman receives a coupon.

I hope that works for you!

3 0

2 years ago

The seventh-grade class is building target areas for a PE activity. The bases for the game will be circular in shape. The diamet

exis [7]

Area of a circle is πr^2.

From this, we can calculate
π*2.5^2
= π*6.25
= 19.625
= 19.6

3 0

2 years ago

Read 2 more answers

- 4 x + 9 = 21 x=___ help

faust18 [17]

X=-3
because
-4x=21-9
-4x= 12 divide both sides
x=-3

4 0

2 years ago

Read 2 more answers

Simplify the following expression as much as you can use exponential properties. (6^-2)(3^-3)(3*6)^4

gtnhenbr [62]

Answer:

Simplifying the expression $(6^{-2})(3^{-3})(3*6)^4$ we get $\mathbf{108}$

Step-by-step explanation:

We need to simplify the expression $(6^{-2})(3^{-3})(3*6)^4$

Solving:

$(6^{-2})(3^{-3})(3*6)^4$

Applying exponent rule: $a^{-m}=\frac{1}{a^m}$

$=\frac{1}{(6^{2})}\frac{1}{(3^{3})}(18)^4\\=\frac{(18)^4}{6^{2}\:.\:3^{3}} \\$

Factors of $18=2\times 3\times 3=2\times3^2$

Factors of $6=2\times 3$

Replacing terms with factors

$=\frac{(2\times3^2)^4}{(2\times 3)^{2}\:.\:3^{3}} \\=\frac{(2)^4\times(3^2)^4}{(2)^2\times (3)^{2}\:.\:3^{3}} \\$

Using exponent rule: $(a^m)^n=a^{m\times n}$

$=\frac{(2)^4\times(3)^8}{(2)^2\times (3)^{2}\:.\:3^{3}} \\=\frac{2^4\times 3^8}{2^2\times 3^{2}\:.\:3^{3}}$

Using exponent rule: $a^m.a^n=a^{m+n}$

$=\frac{2^4\times 3^8}{2^2\times 3^{2+3}}\\=\frac{2^4\times 3^8}{2^2\times 3^{5}}$

Now using exponent rule: $\frac{a^m}{a^n}=a^{m-n}$

$=2^{4-2}\times 3^{8-5}\\=2^{2}\times 3^{3}\\=4\times 27\\=108$

So, simplifying the expression $(6^{-2})(3^{-3})(3*6)^4$ we get $\mathbf{108}$

7 0

2 years ago

A special deck of cards has ten cards. Four are green (G), four are blue (B), and two are red (R). When a card is picked, the co

a_sh-v [17]

Answer:

{GT, GH, BT, BH, RT, RH};

1 /5;

Mutually exclusive ;

Not mutually exclusive.

Step-by-step explanation:

Given :

Green cards, G = 4

Blue cards, B = 4

Red cards, R = 2

For a coin toss :

{H, T}

A card is picked, then a coin is tossed :

Sample space :

{GT, GH, BT, BH, RT, RH}

2.) probability of picking a green card, then probability of landing a head on a coin toss

P(A) = number of required outcome / Total possible outcomes

P(A) = P(green card) * P(Head)

P(A) = (4 / 10) * (1/2)

P(A) = 4 /20

P(A) = 1/5

C.)

Both A and B are mutually exclusive, since both event A and event B cannot occur together, since both red and green cannot be picked during a single pick, this either a red is picked or green is picked, then they are A and B are mutually exclusive.

D.) Event A and C are not mutually exclusive, picking a green card, event A and picking a red or blue card, event B. Both event can happen simultaneously, hence, event A and B are not mutually exclusive.

6 0

2 years ago