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3 years ago

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In one week, each cat at a cat shelter eats 7 cans of cat food. How many cans of cat food are needed for 7 cats eat in one week?

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2 answers:

Shalnov [3]3 years ago

8 0

7+7+7+7+7+7+7=49

#49 cans are needed for 7 cats eat in one week

Llana [10]3 years ago

4 0

49 is your answer 7 x 7 = 49

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Simplify the expression.<br><br><br> Write your answer without negative exponents.

Phoenix [80]

Answer:

-5/ (x^19 * y^12)

Step-by-step explanation:

-15x^-10 * y ^ -3 * x^-9 * y^-9/3 (It's writing it like this.)

15 and 3 have a GCF of 3.

-5x^-10 * y^-3 * x^-9 * y^-9/1

Reduce similar occurrences. *twice in a row*

-5x^(-10 + (-9)) * y^(-3 + (-9))/1

Remove extra signs *twice in a row*

-5x^(-10 - 9) * y^(-3 - 9)/1

Remove extraneous 1.

-5x^(-10 - 9) * y^(-3 - 9)

Add integers.

-5x^-19 * y^-12

Negative powers change multiplication to division.

-5 / (x^19 * y^12) is the final answer.

Even with that, the simplified answer wouldn't have negative exponents because of the negative powers rule.

8 0

2 years ago

Please could someone explain this to me?

qwelly [4]

Ok so A is the same as the square root of 4b+c but all of the numbers and the answer have to be prime numbers

4 0

3 years ago

(2+5xyz-6tv²c)-(11-5yxz-2ctv²)

Effectus [21]

Answer:

-4ctv² + 10xyz - 9

Step-by-step explanation:

(2 + 5xyz - 6tv²c) - (11 - 5yxz - 2ctv²)

2 + 5xyz - 6tv²c - 11 + 5yxz + 2ctv²

-4ctv² + 10xyz - 9

I hope this helps!

5 0

2 years ago

Which of the following expressions is equal to 1?

Vikki [24]

I think it's d okay maybe

8 0

3 years ago

Two athletic teams play a series of games; the first team to win 4 games is declared the overall winner. Suppose that one of the

kodGreya [7K]

Answer:

p=0.7103 (4-game series)
p=0.6480 (2-game series)

Step-by-step explanation:

Let X be the random variable equal the the first 4 straight wins. An overall win for the stronger team implies a negative binomial function with the parameters n=4, p=0.6:

$P(X=4)={{i-1}\choose {4-1}}0.6^40.4^{i-4},\ i=4,5,6,7$

#We find probabilities for the different values of i:

$P(X=4)={3\choose 3}0.6^4=0.1296\\\\P(X=5)={4\choose 3}0.6^40.4^1=0.2074\\\\P(X=6)={5\choose 3}0.6^40.4^2=0.2074\\\\P(X=4)={6\choose 3}0.6^40.4^3=0.1659$

Hence, probability of the stronger team winning overall is:

$=P(X=4)+P(X=5)+P(X=6)+P(X=7)\\\\=0.7103$

#Define Y as the random variable for winning 2/3 games.:

$P(Y=2)={1\choose 1}0.6^2=0.3600\\\\P(Y=3)={2\choose3}0.6^20.4=0.2880\\\\P(win)=0.2880+0.3600=0.6480$

Hence, probability of the stronger team winning in 2 out 3 game series is 0.6480

The stronger team has a higher chance of winning in a 4-game series(0.7103>0.6480)

8 0

3 years ago