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Alex_Xolod [135]
3 years ago
8

C=48y+0.99n solve equation for y

Mathematics
1 answer:
snow_lady [41]3 years ago
4 0
c = 48y + 0.99n\\\\\sf Subtract ~0.99n ~on ~both ~sides\\\\c -0.99n= 48y +0.99n -0.99n\\c -0.99n = 48y\\\\Divide~ 48 ~on~ both~ sides\\\\ \frac{c-0.99n}{48} = \frac{48y}{48} \\\\\boxed {y= \frac{c-0.99n}{48} }
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denis-greek [22]

Answer:

y= (2-√3x+√3)(2+√3x-√3)

Step-by-step explanation:

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use identity : a²-b²=(a-b)(a+b)

y=-3(x-1)^2+4  = 2²- (√3(x-1))²= 2²-(√3x-√3)² = (2-√3x+√3)(2+√3x-√3)

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Refer to the random sample of customer order totals with an average of $78.25 and a population standard deviation of $22.50. a.
zysi [14]

Answer:

a) 78.25- 1.64 \frac{22.50}{\sqrt{40}}= 72.416

78.25+ 1.64 \frac{22.50}{\sqrt{40}}= 84.084

b) 78.25- 1.64 \frac{22.50}{\sqrt{75}}= 73.989

78.25+ 1.64 \frac{22.50}{\sqrt{75}}= 82.511

c) For this case when we increase the sample size the margin of error would be lower and then the interval would be narrower

d)   ME=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}    (a)

Solving for n we got:

n=(\frac{z_{\alpha/2} \sigma}{ME})^2   (b)

And replacing the info we have:

n=(\frac{1.640(22.50)}{5})^2 =54.46 \approx 55

Step-by-step explanation:

Part a

For this case we have the following data given

\bar X = 78.25 represent the sample mean for the customer order totals

\sigma =22.50 represent the population deviation

n= 40 represent the sample size selected

The confidence level is 90% or 0.90 and the significance level would be \alpha=0.1 and \alpha/2 = 0.05 and the critical value from the normal standard distirbution would be given by:

z_{\alpha/2}=1.64

And the confidence interval is given by:

\bar X -z_{\alpha/2} \frac{\sigma}{\sqrt{n}}

And replacing we got:

78.25- 1.64 \frac{22.50}{\sqrt{40}}= 72.416

78.25+ 1.64 \frac{22.50}{\sqrt{40}}= 84.084

Part b

The sample size is now n = 75, but the same confidence so the new interval would be:

78.25- 1.64 \frac{22.50}{\sqrt{75}}= 73.989

78.25+ 1.64 \frac{22.50}{\sqrt{75}}= 82.511

Part c

For this case when we increase the sample size the margin of error would be lower and then the interval would be narrower

Part d

The margin of error is given by:

 ME=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}    (a)

Solving for n we got:

n=(\frac{z_{\alpha/2} \sigma}{ME})^2   (b)

And replacing the info we have:

n=(\frac{1.640(22.50)}{5})^2 =54.46 \approx 55

3 0
3 years ago
Donna bought 5 bags of dog treats for 12.55. What is the cost per bag
Lera25 [3.4K]

Answer:

$2.51

Step-by-step explanation:

12.55/5= 2.51

a bag of dog treats is 2.51 per bag

4 0
2 years ago
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