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muminat
3 years ago
6

A random sample of 12 graduates of a certain secretarial school typed an average of 79.3 words per minute with a standard deviat

ion of 7.8 words per minute. assuming a normal distribution for the number of words typed per minute, find a 95% confidence interval for the average number of words typed by all graduates of this school.
Mathematics
1 answer:
Crazy boy [7]3 years ago
4 0
We will use the following  formula to work out the confidence interval

Upper limit = μ + z* (σ/√n)
Lower limit = μ - z* (σ/√n)

We have
μ = 79.3
σ = 7.8
n = 12
z* is the z-score for 95% confidence level = 1.96

Substitute these into the formula, we have

Upper limit = 79.3 + 1.96 (7.8/√12) = 83.7
Lower limit = 79.3 - 1.96 )7.8/√12) = 74.9
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The expression 4x/5-2x/3 is equivalent to what?
Vinvika [58]

Answer:

(2 x)/15

Step-by-step explanation:

Simplify the following:

(4 x)/5 - (2 x)/3

Put each term in (4 x)/5 - (2 x)/3 over the common denominator 15: (4 x)/5 - (2 x)/3 = (12 x)/15 - (10 x)/15:

(12 x)/15 - (10 x)/15

(12 x)/15 - (10 x)/15 = (12 x - 10 x)/15:

(12 x - 10 x)/15

12 x - 10 x = 2 x:

Answer: (2 x)/15

7 0
3 years ago
(3.2 x 10^14) (5.1 x 10^2) in scientific notation
tresset_1 [31]
Those expressions are already in scientific notation.
6 0
3 years ago
What is 3/4 ÷ 1/6? Please help !......
Reika [66]

Answer:

4 1/2

Step-by-step explanation:

Apply the fractions formula for division, to

3/4÷1/6

and solve

3/6x4/1

=18/4

Reduce by dividing both the numerator and denominator by the Greatest Common Factor GCF(18,4) = 2

18/24÷2/2=9/2

Convert to a mixed number using

long division for 9 ÷ 2 = 4R1, so

9/2=4 1/2

Therefore:

3/4÷1/6=4 1/2

4 0
3 years ago
the Andersons kept track of the time they spent watching television. Part A wich bar diagram represents the number of hours the
DENIUS [597]

Part A:

Part B:

\begin{gathered} 4\frac{5}{6}=\frac{29}{6} \\ 3\frac{4}{6}=\frac{11}{3} \\ so\colon \\ \frac{29}{6}+\frac{11}{3}=\frac{17}{2}=8\frac{3}{6} \end{gathered}

Part C:

\begin{gathered} 3\frac{1}{6}=\frac{19}{6} \\ 2\frac{3}{6}=\frac{15}{6} \\ so\colon \\ \frac{29}{6}+\frac{11}{3}+\frac{19}{6}+\frac{15}{6}=\frac{85}{6}=14\frac{1}{6} \end{gathered}

Part D:

\frac{17}{2}-\frac{17}{3}=\frac{17}{6}=2\frac{5}{6}_{}

4 0
1 year ago
A recent study examined hearing loss data for 1981 U.S. teenagers. In this sample, 369 were found to have some level of hearing
ruslelena [56]

Answer:

There is no enough evidence to support the claim that the proportion of US teens that have some level of hearing loss differs from 20%.

P-value=0.12

Step-by-step explanation:

We have to perform a test of hypothesis on the proportion.

The claim is that the proportion of US teens that have some level of hearing loss differs from 20%.

Then, the null and alternative hypothesis are:

H_0: \pi=0.20\\\\H_a:\pi\neq0.20

The significance level is assumed to be 0.05.

The sample, of size n=1981, has 369 positive cases. Then, the proportion is:

p=X/n=369/1981=0.186

The standard error of the proportion is:

\sigma_p=\sqrt{\dfrac{\pi(1-\pi)}{n}}=\sqrt{\dfrac{0.2*0.8}{1981}}=\sqrt{ 0.000081 }= 0.009

Now, we can calculate the statistic z:

z=\dfrac{p-\pi+0.5/n}{\sigma_p}=\dfrac{0.186-0.20+0.5/1981}{0.009}=\dfrac{-0.014}{0.009}=-1.556

The P-value for this two-tailed test is:

P-value=2*P(z

The P-value is below the significance level, so the effect is not significant. The null hypothesis failed to be rejected.

There is no enough evidence to support the claim that the proportion of US teens that have some level of hearing loss differs from 20%.

7 0
3 years ago
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