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gayaneshka [121]
3 years ago
7

(x+42)3x. solve for x

Mathematics
1 answer:
pishuonlain [190]3 years ago
3 0

Answer: X is equal to either 0 or -42

Step-by-step explanation:

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A yearbook printer charges based on the number of pages printed. Here is a table that shows the cost of some recent yearbooks:
asambeis [7]
y = 0.012x + 0.65<span><span><span>

Number of pages </span> <span>     50.00    </span><span>  100.00 </span><span>   150.00 </span> <span>   200.00 </span> </span> <span> <span> 
Cost </span> <span>                           1.25 </span> <span>         1.85 </span> <span>       2.45 </span> <span>       3.05

x = number of pages
y = cost

intervals of x = 50 pages
intervals of y = 0.60
0.60 / 50 = 0.012

y = 0.012x + 0.65

</span></span></span><span> <span> </span><span><span> <span>                      x </span> <span>      0.012*x </span> <span>              y  </span>
</span> <span> 0.012 <span>           50 </span> <span>         0.60 </span> <span>    0.65 </span> <span>   1.25
</span></span><span>0.012 <span>         100 </span> <span>         1.20 </span> <span>    0.65 </span> <span>   1.85
</span> </span> <span> 0.012 <span>         150 </span> <span>         1.80 </span> <span>    0.65 </span> <span>   2.45
</span> </span> <span> 0.012 <span>         200 </span> <span>         2.40 </span> <span>    0.65 </span> <span>   3.05 </span> </span></span></span><span>
</span>
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3 years ago
Jasmine loves to babysit. On Saturday, she babysat from 3:29 p.m. to 4:52 p.m. On Sunday, she babysat from 1:37 p.m. to 3:16 p.m
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In Sunday she babysits 1 hour and 23 minutes,in Saturday she babysit 1 hour and 39 minutes.In total it’s 3 hours and 12 minutes.
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2 years ago
Suppose the sediment density (g/cm) of a randomly selected specimen from a certain region is normally distributed with mean 2.65
erma4kov [3.2K]

Answer:

Probability that the sample average is at most 3.00 = 0.98030

Probability that the sample average is between 2.65 and 3.00 = 0.4803

Step-by-step explanation:

We are given that the sediment density (g/cm) of a randomly selected specimen from a certain region is normally distributed with mean 2.65 and standard deviation 0.85.

Also, a random sample of 25 specimens is selected.

Let X bar = Sample average sediment density

The z score probability distribution for sample average is given by;

               Z = \frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } } ~ N(0,1)

where, \mu = population mean = 2.65

           \sigma  = standard deviation = 0.85

            n = sample size = 25

(a) Probability that the sample average sediment density is at most 3.00 is given by = P( X bar <= 3.00)

    P(X bar <= 3) = P( \frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } } <= \frac{3-2.65}{\frac{0.85}{\sqrt{25} } } ) = P(Z <= 2.06) = 0.98030

(b) Probability that sample average sediment density is between 2.65 and 3.00 is given by = P(2.65 < X bar < 3.00) = P(X bar < 3) - P(X bar <= 2.65)

P(X bar < 3) = P( \frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } } < \frac{3-2.65}{\frac{0.85}{\sqrt{25} } } ) = P(Z < 2.06) = 0.98030

 P(X bar <= 2.65) = P( \frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } } <= \frac{2.65-2.65}{\frac{0.85}{\sqrt{25} } } ) = P(Z <= 0) = 0.5

Therefore, P(2.65 < X bar < 3)  = 0.98030 - 0.5 = 0.4803 .

                                                                             

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A.29 degrees warmer I hope this helps you
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I hope this helps you

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