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weqwewe [10]
3 years ago
12

Regina earns $9 per hour Jake earns $7.80 per hour they both worked for 8 hours how much money did Regina earn?

Mathematics
2 answers:
Serga [27]3 years ago
8 0
Regina earned $72 you find that out by 9x8
yanalaym [24]3 years ago
5 0
She earned 72 dollars
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How do you find the x- and y-intercepts of an equation?
jeyben [28]

Given two equations containing x and y you are able to find two solutions x and y if they exist.

Hope this helps.

4 0
3 years ago
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The sum of the reciprocals of two consecutive odd integers is 24/143. find the two integers.
nordsb [41]
Let one integer be x, the consecutive odd integer is x+2
reciprocals are 1/x and 1/(x+2)
1/x +1/(x+2)=24/143
make the denominators the same:
(x+2)/[x(x+2)] +x/[x(x+2)]=24/143
143(2x+2)=24(x)(x+2)
286x+286=24x²+48x
24x²-238x-286=0
use the quadratic formula: x=22 this doesn't work because 22 is not an odd number.
or x=-2.1666666666 (this is not an integer)

weird.
4 0
3 years ago
WILL MARK YOU IF YOU CAN HELP plz help me understand this to or add one of my socials to help me with geometry:(
ioda

Answer:

17x + 2 = 16x+6

17x-16x= 6-2

x=4

5 0
3 years ago
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Use the polynomial identity to determine which values of x and y generate the values of the sides of the following right triangl
Dominik [7]
X^2 = 73 - y^2 


<span>x^2 - y^2 = 55 </span>
<span>substituting in to it </span>
<span>73 - 55 = 2y^2 </span>
<span>18 = 2y^2 </span>
<span>y = 3 </span>
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<span>answer is A</span>
8 0
3 years ago
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The lengths of pregnancies in a small rural village are normally distributed with a mean of 262 days and a standard deviation of
RideAnS [48]

Answer:

The range in which we can expect to find the middle 68% of most pregnancies is [245 days , 279 days].

Step-by-step explanation:

We are given that the lengths of pregnancies in a small rural village are normally distributed with a mean of 262 days and a standard deviation of 17 days.

Let X = <u><em>lengths of pregnancies in a small rural village</em></u>

SO, X ~ Normal(\mu=262,\sigma^{2} = 17^{2})

Here, \mu = population mean = 262 days

         \sigma = standard deviation = 17 days

<u>Now, the 68-95-99.7 rule states that;</u>

  • 68% of the data values lies within one standard deviation points.
  • 95% of the data values lies within two standard deviation points.
  • 99.7% of the data values lies within three standard deviation points.

So, middle 68% of most pregnancies is represented through the range of within one standard deviation points, that is;

[ \mu -\sigma , \mu + \sigma ]  =  [262 - 17 , 262 + 17]

                          =  [245 days , 279 days]

Hence, the range in which we can expect to find the middle 68% of most pregnancies is [245 days , 279 days].

5 0
3 years ago
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