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2 years ago

Five cats each ate 1/4 cup of canned food . How much food did they eat altogether?

Mathematics

1 answer:

Sauron [17]2 years ago

4 0

We have FIVE cats that each ate 1/4 cup of canned food.

Then, the five of them together ate:

5 times 1/4 cup This in mathematical expression is a profduct of an integer value (5) times the fraction (1/4) which we know how to multiply based on the problems we solved a few minutes ago:

5 * 1/4 = 5/1 * 1/4 = (5 * 1) / (1 * 4) = 5/4 cups

This answer can also be written as a MIXED number given that 5/4 is an IMPROPER fraction (a fraction with numerator larger than the denominator)

5/4 = 1 1/4

Not sure what answer your teacher would prefer. It depends on whether he/she wants you to use fractions or mixed numbers.

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Which statement compares the median wind speeds for the data in the two box plots? The median wind speed for country A is greate

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How many packages of diapers can you buy with 40$ if one package costs $8?

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4 years ago

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6. Suppose that a fair coin is tossed 2 times, and the result of each toss (H or T) is recorded.

nekit [7.7K]

Answer:

a) $S= {HH, HT, TH, TT}$

b) $P(X=0) = (2C0) (0.5)^0 (1-0.5)^{2-0}= 0.25$

$P(X=1) = (2C1) (0.5)^1 (1-0.5)^{2-1}= 0.5$

$P(X=2) = (2C2) (0.5)^2 (1-0.5)^{2-2}= 0.25$

And we have the following table:

X | 0 | 1 | 2

P(X) | 0.25 | 0.5 | 0.25

Step-by-step explanation:

Let's define first some notation

H= represent a head for the coin tossed

T= represent tails for the coin tossed

We are going to toss a coin 2 times so then the size of the sample size is $2^2 = 4$

a. What is the sample space for this chance experiment?

The sampling space on this case is given by:

$S= {HH, HT, TH, TT}$

b. For this chance experiment, give the probability distribution for the random variable of the total number of heads observed.

The possible values for the number of heads are X=0,1,2. If we assume a fair coin then the probability of obtain heads is the same probability of obtain tails and we can find the distribution like this:

$P(X=0) = (2C0) (0.5)^0 (1-0.5)^{2-0}= 0.25$

$P(X=1) = (2C1) (0.5)^1 (1-0.5)^{2-1}= 0.5$

$P(X=2) = (2C2) (0.5)^2 (1-0.5)^{2-2}= 0.25$

And we have the following table:

X | 0 | 1 | 2

P(X) | 0.25 | 0.5 | 0.25

5 0

3 years ago

Plz solve if you can. Show ur work. I will name Brainliest! (: I need help plzzzzzzzzzz, especially on the check your work part.

maksim [4K]

1) -x + 4 = -2x - 6

Add(2x)

x+4=-6

Subtract(4)

x = -10

Check your work

-(-10)+4=-2(-10) - 6

10+4=-2(-10) - 6

14=-2(-10)

14=20 - 6

14=14

Correct :)

2) 4R - 4 = 3R + 10

Add 4

4R=3R+14

Subtract 3R

R = 14

Check your work

4(14)-4=3(14)+10

56-4=3(14)+10

52=3(14)+10

52=42+10

52=52

Correct :)

3) 2Y - 3 = Y - 4

Add 3

2Y = Y - 1

Subtract Y

Y = -1

Check your work

2(-1)-3=-1-4

-2-3=-1-4

-5=-5

Correct :)

Hope it helps <3

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3 years ago