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Mekhanik [1.2K]
3 years ago
7

By the time shana had completed 3/8 of her first lap in a race, she had also completed 1/32 of the entire race. How many laps we

re there in the race.
Mathematics
1 answer:
Yuri [45]3 years ago
3 0
3/8 of a lap is 1/32 of the entire race.
x - the number of laps in the race

\frac{3}{8}=\frac{1}{32}x \\
\frac{3}{8} \times 32=x \\
3 \times 4=x \\
x=12

There were 12 laps in the race.
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Building and reduce<br><br>1/2 = / 10​
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Answer: 5/10

Explanation: Compare the denominators. Since 2 ⋅ 5 = 10, multiply the numerator 1 by 5 to get 5.

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Step-by-step explanation:

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Step-by-step explanation:

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8 0
3 years ago
If a2= 18 and a4= 8, find two possible common ratios. Then, write two sequences lhaving five terms each.
Marizza181 [45]

common ratio = a4/a3 = a3/a2

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3 0
3 years ago
You are given the sample mean and the population standard deviation. Use this information to construct the​ 90% and​ 95% confide
salantis [7]

Answer:

With​ 90% confidence, it can be said that the population mean price lies in the first interval. With​ 95% confidence, it can be said that the population mean price lies in the second interval. The​ 95% confidence interval is wider than the​ 90%.

Step-by-step explanation:

We are given that a random sample of 60 home theater systems has a mean price of​$131.00. Assume the population standard deviation is​$18.80.

  • Firstly, the pivotal quantity for 90% confidence interval for the  population mean is given by;

                            P.Q. = \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }  ~ N(0,1)

where, \bar X = sample mean price = $131

            \sigma = population standard deviation = $18.80

            n = sample of home theater = 60

            \mu = population mean

<em>Here for constructing 90% confidence interval we have used One-sample z test statistics as we know about the population standard deviation.</em>

<u>So, 90% confidence interval for the population mean, </u>\mu<u> is ;</u>

P(-1.645 < N(0,1) < 1.645) = 0.90  {As the critical value of z at 5% level

                                                   of significance are -1.645 & 1.645}  

P(-1.645 < \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } } < 1.645) = 0.90

P( -1.645 \times {\frac{\sigma}{\sqrt{n} } } < {\bar X-\mu} < 1.645 \times {\frac{\sigma}{\sqrt{n} } } ) = 0.90

P( \bar X-1.645 \times {\frac{\sigma}{\sqrt{n} } } < \mu < \bar X+1.645 \times {\frac{\sigma}{\sqrt{n} } } ) = 0.90

<u>90% confidence interval for</u> \mu = [ \bar X-1.645 \times {\frac{\sigma}{\sqrt{n} } } , \bar X+1.645 \times {\frac{\sigma}{\sqrt{n} } } ]

                                                  = [131-1.645 \times {\frac{18.8}{\sqrt{60} } } , 131+1.645 \times {\frac{18.8}{\sqrt{60} } } ]

                                                  = [127.01 , 134.99]

Therefore, 90% confidence interval for the population mean is [127.01 , 134.99].

  • Now, the pivotal quantity for 95% confidence interval for the  population mean is given by;

                            P.Q. = \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }  ~ N(0,1)

where, \bar X = sample mean price = $131

            \sigma = population standard deviation = $18.80

            n = sample of home theater = 60

            \mu = population mean

<em>Here for constructing 95% confidence interval we have used One-sample z test statistics as we know about the population standard deviation.</em>

<u>So, 95% confidence interval for the population mean, </u>\mu<u> is ;</u>

P(-1.96 < N(0,1) < 1.96) = 0.95  {As the critical value of z at 2.5% level

                                                   of significance are -1.96 & 1.96}  

P(-1.96 < \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } } < 1.96) = 0.95

P( -1.96 \times {\frac{\sigma}{\sqrt{n} } } < {\bar X-\mu} < 1.96 \times {\frac{\sigma}{\sqrt{n} } } ) = 0.95

P( \bar X-1.96 \times {\frac{\sigma}{\sqrt{n} } } < \mu < \bar X+1.96 \times {\frac{\sigma}{\sqrt{n} } } ) = 0.95

<u>95% confidence interval for</u> \mu = [ \bar X-1.96 \times {\frac{\sigma}{\sqrt{n} } } , \bar X+1.96 \times {\frac{\sigma}{\sqrt{n} } } ]

                                                  = [131-1.96 \times {\frac{18.8}{\sqrt{60} } } , 131+1.96 \times {\frac{18.8}{\sqrt{60} } } ]

                                                  = [126.24 , 135.76]

Therefore, 95% confidence interval for the population mean is [126.24 , 135.76].

Now, with​ 90% confidence, it can be said that the population mean price lies in the first interval. With​ 95% confidence, it can be said that the population mean price lies in the second interval. The ​95% confidence interval is wider than the​ 90%.

7 0
3 years ago
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