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Romashka-Z-Leto [24]
3 years ago
7

The amount of money spent weekly on cleaning, maintenance, and repairs at a large restaurant was observed over a long period of

time to be approximately normally distributed, with mean μ = $597 and standard deviation σ = $39.
(a) If $646 is budgeted for next week, what is the probability that the actual costs will exceed the budgeted amount?
Mathematics
1 answer:
Kazeer [188]3 years ago
7 0
Z = ( X - Mean ) / SD
z = ( 646 - 597 ) / 39
z = 49/39 = 1.25
With the z-score table:
for z = 1.25, p = 0.8944 ( probability that the cost will be lower than $646 )
1 - 0.8944 = 0.1056
Answer: <span>The probabilit</span>y is 0.1056
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Simplify the following:

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3 0
3 years ago
Write an exponential function in the form y=ab^xy=ab
Zarrin [17]

Answer:

Step-by-step explanation:

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8 0
2 years ago
Prove that the diagonals of a parallelogram bisect each other​
Nady [450]

Answer:

[ See the attached picture ]

The diagonals of a parallelogram bisect each other.

✧ Given : ABCD is a parallelogram. Diagonals AC and BD intersect at O.

✺ To prove : AC and BD bisect each other at O , i.e AO = OC and BO = OD.

Proof :\begin{array}{ |c| c |  c |  } \hline \tt{SN}& \tt{STATEMENTS} & \tt{REASONS}\\ \hline 1& \sf{In  \: \triangle ^{s}  \:AOB \: and \: COD  } \\  \sf{(i)}&  \sf{ \angle \: OAB =  \angle \: OCD\: (A)}& \sf{AB \parallel \: DC \: and \: alternate \: angles} \\  \sf{(ii)} &\sf{AB = DC(S)}& \sf{Opposite \: sides \: of \: a \: parallelogram} \\  \sf{(iii)} &\sf{ \angle \: OBA=  \angle \: ODC(A)} &\sf{AB \parallel \:DC \: and \: alternate \: angles} \\  \sf{(iv)}& \sf{ \triangle \:AOB\cong \triangle \: COD}& \sf{A.S.A \: axiom}\\ \hline 2.& \sf{AO = OC \: and \: BO = OD}& \sf{Corresponding \: sides \: of \: congruent \: triangle}\\ \hline 3.& \sf{AC \: and \: BD \: bisect \: each \: other \: at \: O}& \sf{From \: statement \: (2)}\\ \\ \hline\end{array}.          Proved ✔

♕ And we're done! Hurrayyy! ;)

# STUDY HARD! So, Tomorrow you can answer people like this , " Dude , I just bought this expensive mobile phone but it is not that expensive for me" [ - Unknown ] :P

☄ Hope I helped! ♡

☃ Let me know if you have any questions! ♪

\underbrace{ \overbrace  {\mathfrak{Carry \: On \: Learning}}} ☂

▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁

5 0
3 years ago
This is an absolute value problem with LETTERS?! Absolute madness (no pun intended)
MAVERICK [17]

Answer:

a

Step-by-step explanation:

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8 0
3 years ago
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