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Lady_Fox [76]
2 years ago
13

If staff salaries were $32,000/month last year and $47,000/month this year, what is the percentage labor cost increase? (Round t

o the nearest percent.)
Mathematics
1 answer:
NeX [460]2 years ago
6 0
100(47000-32000)/32000=47%  (to nearest percent)
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For two days in a row Alvin jogged in the park he jogged for 46 min on Friday this is 19 min less than he ran on Saturday write
Luda [366]
Saturday = 46+19 min = 65 minutes
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2 years ago
The center of a hyperbola is located at (0, 0). One focus is located at (0, 5) and its associated directrix is represented by th
Lady bird [3.3K]
For a hyperbola   \dfrac{y^{2}}{a^{2}}-\dfrac{x^{2}}{b^{2}}=1
where   a^{2}+b^{2}=c^{2}
the directrix is the line   y=\dfrac{a^{2}}{c}
and the focus is at (0, c).

Here, we have c = 5, a² = 9, so b² = 5² - 9 = 16.
  a = √9 = 3
  b = √16 = 4

Your hyperbola's constants are ...
  a = 3
  b = 4


______
Please note that the equation of a hyperbola has a negative sign for one of the terms. The equation given in your problem statement is that of an ellipse.
8 0
2 years ago
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Assume that you pay $2,849.84 in state property taxes every year. If your property has an assessed value of $41,302, what is you
Olenka [21]
The answer would be D
4 0
2 years ago
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An automated egg carton loader has a 1% probability of cracking an egg, and a customer will complain if more than one egg per do
vaieri [72.5K]

Answer:

a) Binomial distribution B(n=12,p=0.01)

b) P=0.007

c) P=0.999924

d) P=0.366

Step-by-step explanation:

a) The distribution of cracked eggs per dozen should be a binomial distribution B(12,0.01), as it can model 12 independent events.

b) To calculate the probability of having a carton of dozen eggs with more than one cracked egg, we will first calculate the probabilities of having zero or one cracked egg.

P(k=0)=\binom{12}{0}p^0(1-p)^{12}=1*1*0.99^{12}=1*0.886=0.886\\\\P(k=1)=\binom{12}{1}p^1(1-p)^{11}=12*0.01*0.99^{11}=12*0.01*0.895=0.107

Then,

P(k>1)=1-(P(k=0)+P(k=1))=1-(0.886+0.107)=1-0.993=0.007

c) In this case, the distribution is B(1200,0.01)

P(k=0)=\binom{1200}{0}p^0(1-p)^{12}=1*1*0.99^{1200}=1* 0.000006 = 0.000006 \\\\ P(k=1)=\binom{1200}{1}p^1(1-p)^{1199}=1200*0.01*0.99^{1199}=1200*0.01* 0.000006 \\\\P(k=1)= 0.00007\\\\\\P(k\leq1)=0.000006+0.000070=0.000076\\\\\\P(k>1)=1-P(k\leq 1)=1-0.000076=0.999924

d) In this case, the distribution is B(100,0.01)

We can calculate this probability as the probability of having 0 cracked eggs in a batch of 100 eggs.

P(k=0)=\binom{100}{0}p^0(1-p)^{100}=0.99^{100}=0.366

5 0
2 years ago
Mary spent $45 at the mall. She bought lunch for $9. She bought 3 shirts at
oee [108]

Answer:

x = 12

Step-by-step explanation:

If Mary spent $45 altogether,

then she bought lunch for $9

so our equation now is

$45 = $9 + 3x

It is 3x because he bought 3 shirts and we used x because we didnt

now how much the price of those shirts were

$45 = $9 +3x

$45 - $9 = 3x

$45 - $9 = $36

$36 = 3x

$36 / 3 = 12

x = 12

Please mark this answer as the brainliest

4 0
2 years ago
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