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2 years ago

I've given up on math, help would be wildly appreciated

Mathematics

1 answer:

REY [17]2 years ago

3 0

Answer:

1. $\frac{2}{3}$ probability

2. head represents cats (let's say) and tails represent dogs, flip two coins to determine whether the family has 2 cats or 2 dogs or 1 cat and 1 dog.

3. Take two coins, flip them, record heads and tails of each flip for 50 flips (I ain't no doing this)

4 - 6. answer these questions based on the data you got...

The End

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5z^3 - 2z^4 - 9z^2 + z

likoan [24]

Answer:

-z( 2z^3-5z^2+9z-1)

Step-by-step explanation:

Apply exponent rule: a^b+c=a^b*a^c

z^2=zz

z^3=z^2*z

z^4=z^3*z

Factor out common term z:

-z(2z^3-5z^2+9z-1)

8 0

3 years ago

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PtichkaEL [24]

Answer:

Step-by-step explanation:

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5 0

2 years ago

Please Help Me worth 20 points

rodikova [14]

Answer:

1.Amy should ask every 10th person leaving the school at the end of the day and Amy should ask the 50 students in her computer lab.

2.Yes,the sample is unbiased and equal to everyone.

3.200

Step-by-step explanation:

8 0

3 years ago

Quasim ordered 32 cases with 28 cans of fruit in each case. He also received 24 cases with 13 large cans of fruit in each case.

klasskru [66]

I am assuming the 32 cases contain the small cans...

32 cases...with 28 small cans of fruit in each case..
32 * 28 = 896 small cans of fruit in total

24 cases...with 13 large cans of fruit in each case...
24 * 13 = 312 large cans of fruit in total

so there are (896 - 312) = 584 more small cans then large cans <==

8 0

2 years ago

In a bag of m&m's there are 5 brown 6 yellow 4 blue 3 green and 2 orange. What's the probability of getting 3 yellow m&m

olasank [31]

There are 5+6+4+3+2=20 m&m's in the bag.
Calculate in how many ways you can choose 3 m&m's from 20:
$_{20} C _3=\frac{20!}{3!(20-3)!}=\frac{20!}{3! \times 17!}=\frac{17! \times 18 \times 19 \times 20}{6 \times 17!}=\frac{18 \times 19 \times 20}{6}=3 \times 19 \times 20= \\
=1140$

There are 6 yellow m&m's.
Calculate in how many ways you can choose 3 m&m's from 6:
$_6 C _3 = \frac{6!}{3!(6-3)!}=\frac{6!}{3! \times 3!}=\frac{3! \times 4 \times 5 \times 6}{3! \times 6}=\frac{4 \times 5 \times 6}{6}=4 \times 5=20$

The probability is the number of ways of choosing 3 m&m's from 6 m&m's divided by the number of ways of choosing 3 m&m's from 20 m&m's.
$P=
\frac{20}{1140}=\frac{20 \div 20}{1140 \div 20}=\frac{1}{57}$

The probability is 1/57.

4 0

2 years ago