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3 years ago

PLSSS HELP 50 POINTSS ILL SEND 40 MORE IF U HELPP PLSSS

1 answer:

3 0

1) y=53

2)53

21 is 2 thirds of 32 so you add them together and get 53. Also Y=Cost of last years trip which we determined is $53

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Geometry Angles question 4, Thanks if you help

Nastasia [14]

Answer:

Question 4: (C)

Question 5: (B)

Step-by-step explanation:

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4 years ago

In a 56 pound mixture of sand and gravel, 14 pound is gravel. How much gravel is there in a 1-pound mixture? 445 lb 313 l

Ivenika [448]

Divide the amount of gravel in the 5/6 pound mixture:

1/4 / 5/6 = 1/4 * 6/5 = 6/20 = 3/10 pounds

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3 years ago

What is the sum of the infinite geometric series?

Alexandra [31]

Answer:

-288

Step-by-step explanation:

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4 years ago

Read 2 more answers

PLS HELP QUICK!!

aliya0001 [1]

Answer:

Step-by-step explanation:

750=3.14 × r^2 ×15 you divide 750 by 3.14 then by 15, then you take the square root

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3 years ago

Calculating conditional probabilities - random permutations. About The letters (a, b, c, d, e, f, g) are put in a random order.

Evgesh-ka [11]

A="b is in the middle"

B="c is to the right of b"

C="The letter def occur together in that order"

a) b can be in 7 places, but only one is the middle. So, P(A)=1/7

b) X=i, "b is in the i-th position"

Y=j, "c is in the j-th position"

$P(B)=\displaystyle\sum_{i=1}^{6}(P(X=i)\displaystyle\sum_{j=i+1}^{7}P(Y=j))=\displaystyle\sum_{i=1}^{6}\frac{1}{7}(\displaystyle\sum_{j=i+1}^{7}\frac{1}{6})=\frac{1}{42}\displaystyle\sum_{i=1}^{6}(\displaystyle\sum_{j=i+1}^{7}1)=\frac{6+5+4+3+2+1}{42}=\frac{1}{2}$

P(B)=1/2

c) X=i, "d is in the i-th position"

Y=j, "e is in the j-th position"

Z=k, "f is in the i-th position"

$P(C)=\displaystyle\sum_{i=1}^{5}( P(X=i)P(Y=i+1)P(Z=i+2))=\displaystyle\sum_{i=1}^{5}(\frac{1}{7}\times\frac{1}{6}\times\frac{1}{5})=\frac{1}{210}\displaystyle\sum_{i=1}^{5}(1)=\frac{1}{42}$

P(C)=1/42

P(A∩C)=2*(1/7*1/6*1/5*1/4)=1/420

$P(B\cap C)=\displaystyle\sum_{i=1}^{3} P(X=i)P(Y=i+1)P(Z=i+2)\displaystyle\sum_{j=i+3}^{6}P(V=j)P(W=j+1)=\displaystyle\sum_{i=1}^{3}\frac{1}{6}\frac{1}{7}\frac{1}{5}(\displaystyle\sum_{j=1+3}^{6}\frac{1}{4}\frac{1}{3})=1/420$

P(B∩A)=3*(1/7*1/6)=1/14

P(A|C)=P(A∩C)/P(C)=(1/420)/(1/42)=1/10

P(B|C)=P(B∩C)/P(C)=(1/420)/(1/42)=1/10

P(A|B)=P(B∩A)/P(B)=(1/14)/(1/2)=1/7

P(A∩B)=1/14

P(A)P(B)=(1/7)*(1/2)=1/14

A and B are independent

P(A∩C)=1/420

P(A)P(C)=(1/7)*(1/42)=1/294

A and C aren't independent

P(B∩C)=1/420

P(B)P(C)=(1/2)*(1/42)=1/84

B and C aren't independent

8 0

3 years ago