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Dominik [7]
3 years ago
14

The weight of eggs from a local farm are Normally

Mathematics
1 answer:
SVEN [57.7K]3 years ago
5 0

Answer:

65.5

Step-by-step explanation:

I just took the assignment on edge. Good luck

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wyatt has (2x-1) one-dollar bills and (4x + 2) five dollar bills. Susan has 3 x dollars more than Wyatt. Find the total amount o
uysha [10]

Answer:

a) $ 22x -9

b) 44x - 18

c) $319

Step-by-step explanation:

Given that,

number of $1 bills wyatt has = (2x-1)

number of $5 bills wyatt has = (4x + 2)

The total amount of money that Wyatt has in terms of x

(2x - 1) + 5(4x + 2)

= 2x - 1 + 20x + 10

= $ 22x -9

The number of pens that Wyatt can buy if each pen costs 50 cents

$(22x - 9) * 100

= 2200x - 900 cents / 50

= 44x - 18

If x = 21, amount of money Susan will have now if Wyatt gives her half the number of five dollar bills that he has

( 5 (4x + 2) / 2) + ( 2 (21) - 1)

= 215 + 41

= 256

As we know that Susan has $3x more than Wyatt

So,

256 + 3x

= 256 + 3(21)

= $319

5 0
3 years ago
Read 2 more answers
Verify identities Will give BRAINLIST<br><br> cot^2(x) - csc^2(x) = -1<br><br> show steps
san4es73 [151]

Answer:

Step-by-step explanation:

∫tan(x)dx∫cot(x)dx∫sec(x)dx∫csc(x)dx====−ln∣∣cos(x)∣∣+C=ln∣∣sec(x)∣∣+Cln∣∣sin(x)∣∣+C=−ln∣∣csc(x)∣∣+Cln∣∣sec(x)+tan(x)∣∣+C−ln∣∣csc(x)+cot(x)∣∣+C

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3 years ago
Write a function to give arbitrary probabilities (totalling 100%) to each face of a six-sided die.
Andre45 [30]
P(A) = N/0
where P(A) equals Probability of any event occurring
N is the Number of ways an event can occur and
0 is the total number of possible Outcomes

P(A) = 1/6

Plainly the probability of rolling a six with a single six-sided dice is one event in which it lands with six uppermost, divided by six possible outcomes from a single throw, or one sixth (16.66 per cent).
7 0
2 years ago
Which point on the grid have they NOT explored?
Debora [2.8K]

Answer:

D

Step-by-step explanation:

5 0
2 years ago
Read 2 more answers
1) AU (C - B) =
torisob [31]

Answer:

A \cup (C-B) = (A \cup C) \cap  (A \cup B^c)\\A \cup (B \cap C) = (A \cup B) \cap (A \cup C) \,\,\, \text{Here you use double inclusion.}\\A \cup B \cup C =  A \cup (B \cup C)\\B^c \cap (A-C)  = (B^c - C) \cap A\\(B \cap C) - A = (B-A)\cap C\\C-(A\cup B) =  (C - A )\cap B\\A^c \cap (B \cap C) = ( A^c \cap B) \cap C

Step-by-step explanation:

A \cup (C-B) = A \cup (C \cap B^c) = (A \cup C) \cap  (A \cup B^c)\\A \cup (B \cap C) = (A \cup B) \cap (A \cup C) \,\,\, \text{Here you use double inclusion.}\\A \cup B \cup C = (A \cup B ) \cup C  = A \cup (B \cup C)\\B^c \cap (A-C)  = B^c \cap (A \cap C^c) =  (B^c \cap C^c) \cap A =  (B^c - C) \cap A\\(B \cap C) - A = (B \cap C) \cap A^c  = (B \cap C) \cap A^c = (B \cap A^c)\cap C = (B-A)\cap C\\C-(A\cup B) = C \cap (A\cup B)^c = C \cap (A^c \cap B^c ) = (C  \cap A^c )\cap B^c = (C - A )\cap B\\

A^c \cap (B \cap C) = ( A^c \cap B) \cap C

3 0
3 years ago
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