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Likurg_2 [28]
3 years ago
11

18

Mathematics
1 answer:
Lorico [155]3 years ago
8 0

Answer:

I just found it out, 200 Student tickets were sold.

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Which expresion represents the perimeter of the rectangle above (3y-5), 2y
Sonbull [250]
(3y-5 + 2y)*2 = (5y-5)*2 = 10y-10
3 0
3 years ago
Read 2 more answers
Which equation creates an infinite number of solutions when solved for a system with <img src="https://tex.z-dn.net/?f=y%3D8x-9"
Tatiana [17]

Answer:

d) 4y − 32x = -36

Step-by-step explanation:

If there are an infinite number of solutions, the equations represent the same line.

y = 8x − 9

y − 8x = -9

4y − 32x = -36

6 0
3 years ago
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Express 80 as the product of its prime factors.<br> Write the prime factors in ascending order.
stepan [7]

Answer:

the answer of that is 16 x 5

Step-by-step explanation:

7 0
2 years ago
Chapter 6, Section 1-D, Exercise 009 Is a Normal Distribution Appropriate? In each case below, is the sample size large enough s
Xelga [282]

Answer:

Step-by-step explanation:

Complete Question:

Chapter 6, Section 1-D, Exercise 009 Is a Normal Distribution Appropriate? In each case below, is the sample size large enough so that the sample proportions follow a normal distribution?

a) n=600 p=0.2

b) n=20, p=0.4

if np=10 and npq=10 then  the data follows normal distribution

a) np= 120,

q= 1-0.2= 0.8

npq= 600 ×0.2×0.4 = 48

Normal distribution is appropriate and sample size is large enough

b) np= 8

q= 1-0.4= 0.6

npq= 20 × 0.4×0.6= 4.8

sample size is not large enough so normal distribution is not appropriate.

4 0
3 years ago
A sample of 15 commuters in Chicago showed the average of the commuting times was 33.2 minutes. If s = 8.3 minutes, find the 95%
OleMash [197]

Answer:

The 95% confidence interval of the true mean.

(29.4261 ,36.9739)

Step-by-step explanation:

<u>Step :- (i)</u>

Given sample size 'n' =15

sample of the mean x⁻ = 33.2

The standard deviation of the sample 'S' = 8.3

<u>95% of confidence intervals</u>

<u></u>(x^{-} - t_{\alpha } \frac{S}{\sqrt{n} } ,x^{-} + t_{\alpha }\frac{S}{\sqrt{n} } )<u></u>

<u>Step:-(ii)</u>

<u>The degrees of freedom γ=n-1 = 15-1=14</u>

The tabulated value t = 1.761 at 0.05 level of significance.

now substitute all possible values, we get

(33.2 - 1.761\frac{8.3}{\sqrt{15} } ,33.2+ 1.761\frac{8}{\sqrt{15} } )

After calculation , we get

(33.2-3.7739 , 33.2+3.7739

(29.4261 ,36.9739)

<u>Conclusion</u>:-

the 95% confidence interval of the true mean.

(29.4261 ,36.9739)

8 0
3 years ago
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