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1 year ago

Janet can read 3.6 pages of a book in a minute. If she read for 2.71 minutes. How much would have she had read

Mathematics

1 answer:

Aliun [14]1 year ago

8 0

Answer:

9.76 pages

Step-by-step explanation:

multiply 3.6 times 2 to get 7.2. Then multiply 3.6 times .71 to get 2.56 then add 7.2 and 2.56 to get 9.76 as your answer.

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If a function f(m) models the number of hours it take to hike m miles through the jungle, which describes the appropriate domain

IrinaK [193]

the correct answer is d

3 0

1 year ago

Read 2 more answers

A savings and loan association needs information concerning the checking account balances of its local customers. A random sampl

Citrus2011 [14]

Answer:

a) 98% confidence interval for the true mean checking account balance for local customers.

(453.586 , 874.693)

b) 95% confidence interval for the standard deviation.

(214.91 , 441.53)

Step-by-step explanation:

Given a size of sample 'n' =14

given mean of the sample x⁻ = $664.14

standard deviation of the sample 'S' = $297.29.

98% of confidence intervals

$(x^{-} - t_{\alpha } \frac{S}{\sqrt{n} } , x^{-}+ t_{\alpha }\frac{S}{\sqrt{n} } )$

The degrees of freedom γ=n-1 =14-1 =13

t₁₃ = 2.650 at 98% of confidence level of signification.

$(664.14- 2.650\frac{297.29}{\sqrt{14} } , 664.14+ 2.650\frac{297.29}{\sqrt{14} } )$

on calculation, we get

(664.14-210.553 , 664.14+210.553)

(453.586 , 874.693)

98% confidence interval for the true mean checking account balance for local customers.

(453.586 , 874.693)

95% of confidence intervals

$({s\sqrt{\frac{n-1}{X^{2} _{(\frac{\alpha }{2} ,n-1) } } } ,s\sqrt{\frac{n-1}{X^2_{\frac{1-\alpha }{2},n-1 } } } )$

The degrees of freedom γ=n-1 =14-1 =13

X^2_{0.05,13} =22.36 (check table)

X^2_{0.95,13} = 5.892 (check table)

$(297.29. (\sqrt{\frac{14-1}{X^2_{0.05,13} } } ),297.29(\sqrt{\frac{14-1}{X^2_{0.95,13} } } )$

$(297.29. (\sqrt{\frac{14-1}{22.36 } } ),297.29(\sqrt{\frac{14-1}{5.892 } } )$

(214.91 , 441.53)

95% confidence interval for the standard deviation.

(214.91 , 441.53)

7 0

1 year ago

The extract of a plant native to Taiwan has been tested as a possible treatment for Leukemia. One of the chemical compounds prod

True [87]

Answer:

a) 57.35%

b) 99.99%

c) 68.27%

Step-by-step explanation:

When we have a random variable X that is normally distributed with mean $\large\bf \mu$ and standard deviation $\large\bf \sigma$ , then

The probability that the random variable has a value less than a, P(X < a) = P(X ≤ a) is the area under the normal curve with mean $\large\bf \mu$ and standard deviation $\large\bf \sigma$ to the left of a.

The probability that the random variable has a value greater than b, P(X > b) = P(X ≥ b) is the area under the normal curve with mean $\large\bf \mu$ and standard deviation $\large\bf \sigma$ to the right of b.

The probability that the random variable has a value between a and b, P(a < X < b) = P(a ≤ X ≤ b) = P(a < X ≤ b)= P(a ≤ X < b) is the area under the normal curve with mean $\large\bf \mu$ and standard deviation $\large\bf \sigma$ between a and b.

In this case, the random variable is the collagen amount found in the extract of the plant. The mean is 63 g/ml and the standard deviation is 5.4 g/ml

(a) What is the probability that the amount of collagen is greater than 62 grams per mililiter?

As we have seen, we need to find the area under the normal curve with mean 63 and standard deviation 5.4 to the right of 62 (see picture).

You can find this value easily with a calculator or a spreadsheet. If you prefer the old-style, then you have to standardize the values and look up in a table.

If you have access to Excel or OpenOffice Calc, you can find this value by introducing the formula:

1- NORMDIST(62,63,5.4,1) in Excel

1 - NORMDIST(62;63;5.4;1) in OpenOffice Calc

and we will get a value of 0.5735 or 57.35%

(b) What is the probability that the amount of collagen is less than 90 grams per mililiter?

Now we want the area to the left of 90

NORMDIST(90,63,5.4,1) in Excel

NORMDIST(90;63;5.4;1) in OpenOffice Calc

You will get a value of 0.9999 or 99.99%

(c) What percentage of compounds formed from the extract of this plant fall within 1 standard deviations of the mean?

You can use either the rule that 68.27% of the data falls between $\large\bf \mu -\sigma$ and $\large\bf \mu +\sigma$ or compute area between 63 - 5.4 and 63 + 5.4, that is to say, the area between 57.6 and 68.4