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3 years ago

8

Five toymakers each carved 28 blocksand 17 airplanes. Three other toymakerseach carved the same number ofairplanes and twice as

many blocks. Howmany toys did the eight carve in all?

1 answer:

timofeeve [1]3 years ago

4 0

Answer:

444

Step-by-step explanation:

28*5

17*5

17*3

56*3

add the answers

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Given a standard deck of 52 cards, 3 cards are dealt without replacement. Using this situation, answer the questions below.<b

kherson [118]

Given that <span>3 cards are dealt without replacement in a </span><span>standard deck of 52 cards.

Part A:

There are 4 queens in a standard deck of 52 card, thus the probability that the first card is a queen is given by 4 / 52 = 1 / 13.

Since, the first card is not replaced, thus there are 3 queens remaining and 51 ards remaining in total, thus the probability that the second card is a queen is given</span> by 3 / 51 = 1 / 17

Similarly the probability that the third card is a queen is given by 2 / 50 = 1 / 25.

Therefore, the probability that <span>all three cards are queens is given by

$\frac{1}{13} \times \frac{1}{17} \times \frac{1}{25} = \frac{1}{5525}$

Part B:

Yes the probability of drawing a queen of heart is independent of the probability of drawing a queen of diamonds because they are separate cards and drawing one of the cards does not in any way affect the chance of drawing the other card.

Part C:

Given that the first card is a queen, then there are 3 queens remaining out of 51 cards remaining, thus the number of cards that are not queen is 51 - 3 = 48 cards.

Therefore, </span>if the first card is a queen, the probability that the second card will not be a queen is given by 48 / 51 = 16 / 17

Part D:

<span>Given that the first two card are queens, then there are 2 queens remaining out of 50 cards remaining.

Therefore, </span>if two of the three cards are queens ,<span>the probability that you will be dealt three queens</span> is given by 2 / 50 = 1 / 25 = 0.04

Part E:

<span>Given that the first two card are queens, then there are 2 queens remaining out of 50 cards remaining, thus the number of cards that are not queen is 50 - 2 = 48 cards.

Therefore, </span>if two of the three cards are queens ,the probability that the other card is not a queen is given by 48 / 50 = 24 / 25 = 0.96

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3 years ago

6th Grade End-of-Module Review Part-xwhole Name Adam Robertson Date 1/12/21 1. Jazzlyn is applying for a job online. She has com

elena55 [62]

Answer:

48 questions.

Step-by-step explanation:

x is for the unknown variable, the number you are trying to figure out.

12= 25%x

if you convert to a fraction, 12 is 1/4 of x. if you then multiply your answer by 4, 12 is 25% of 48.

to figure this out you can also ask, 12 is 25% of what number?

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3 years ago

What is the equation of the line that is parallel to y=2x+8yand goes through point (−4,1)?

vekshin1

<span>y=2x+8 slope = 2

</span><span>parallel to y=2x+8, so equation has same slope = 2
</span><span>goes through point (−4,1)
</span>then
y = mx+b
1 = 2(-4) + b
b = 9

so the equation
y = 2x + 9

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3 years ago

If the ratio of boys to girls is 2:3 and there are 15 girls in your class, how many boys are there.

kakasveta [241]

Answer:

10

Step-by-step explanation:

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4 years ago

Read 2 more answers

Determine whether each of the binary relation R defined on the given sets A is reflexive, symmetric, antisymmetric, or transitiv

ki77a [65]

Answer:

In explanation

Please let me know if something doesn't make sense.

Step-by-step explanation:

a)

*This relation is not reflexive.

0 is an integer and (0,0) is not in the relation because 0(0)>0 is not true.

*This relation is symmetric because if a(b)>0 then b(a)>0 since multiplication is commutative.

*This relation is transitive.

Assume a(b)>0 and b(c)>0.

Note: This means not a,b, or c can be zero.

Therefore we have abbc>0.

Since b^2 is positive then ac is positive.

Since a(c)>0, then (a,c) is in R provided (a,b) and (b,c) is in R.

*The relation is not antisymmretric.

(3,2) and (2,3) are in R but 3 doesn't equal 2.

b)

*This relation is reflective.

Since a^2=a^2 for any a, then (a,a) is in R.

*The relation is symmetric.

If a^2=b^2, then b^2=a^2.

*The relation is transitive.

If a^2=b^2 and b^2=c^2, then a^2=c^2.

*The relation is not antisymmretric.

(1,-1) and (-1,1) is in the relation but-1 doesn't equal 1.

c)

*The relation is reflexive.

a/a=1 for any a in the naturals.

*The relation is not symmetric.

Wile 4/2 is an integer, 2/4 is not.

*The relation is transitive.

If a/b=z and b/c=y where z and y are integers, then a=bz and b=cy.

This means a=cyz. This implies a/c=yz.

Since the product of integers is an integer, then (a,c) is in the relation provided (a,b) and (b,c) are in the relation.

*The relation is antisymmretric.

Assume (a,b) is an R. (Note: a,b are natural numbers.) This means a/b is an integer. This also means a is either greater than or equal to b. If b is less than a, then (b,a) is not in R. If a=b, then (b,a) is in R. (Note: b/a=1 since b=a)

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3 years ago