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Andreas93 [3]
2 years ago
9

What shape is this cross-section?

Mathematics
1 answer:
hoa [83]2 years ago
5 0

Answer:

Circle

Step-by-step explanation:

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Step-by-step explanation:

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Which equation can be used to find the side lengths if the longest side measures 6.2cm?
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2 years ago
A study was designed to compare the attitudes of two groups of nursing students towards computers. Group 1 had previously taken
trasher [3.6K]

Answer:

Kindly check explanation

Step-by-step explanation:

Given the following :

Group 1:

μ1 = 59.7

s1 = 2.8

n1 = sample size = 12

Group 2:

μ2 = 64.7

s2 = 8.3

n2 = sample size = 15

α = 0.1

Assume normal distribution and equ sample variance

A.)

Null and alternative hypothesis

Null : μ1 = μ2

Alternative : μ1 < μ2

B.)

USing the t test

Test statistic :

t = (m1 - m2) / S(√1/n1 + 1/n2)

S = √(((n1 - 1)s²1 + (n2 - 1)s²2) / (n1 + n2 - 2))

S = √(((12 - 1)2.8^2 + (15 - 1)8.3^2) / (12 + 15 - 2))

S = 6.4829005

t = (59.7 - 64.7) / 6.4829005(√1/12 + 1/15)

t = - 5 / 2.5108165

tstat = −1.991384

Decision rule :

If tstat < - tα, (n1+n2-2) ; reject the Null

tstat < t0.1,25

From t table :

-t0.1, 25 = - 1.3163

tstat = - 1.9913

-1.9913 < - 1.3163 ; Hence reject the Null

5 0
3 years ago
A survey was conducted that asked 1003 people how many books they had read in the past year. Results indicated that x overbar eq
Sergio [31]

Answer:

The 99% confidence interval would be given (11.448;14.152).

Step-by-step explanation:

1) Important concepts and notation

A confidence interval for a mean "gives us a range of plausible values for the population mean. If a confidence interval does not include a particular value, we can say that it is not likely that the particular value is the true population mean"

s=16.6 represent the sample deviation

\bar X=12.8 represent the sample mean

n =1003 is the sample size selected

Confidence =99% or 0.99

\alpha=1-0.99=0.01 represent the significance level.

2) Solution to the problem

The confidence interval for the mean would be given by this formula

\bar X \pm z_{\alpha/2} \frac{s}{\sqrt{n}}

We can use a z quantile instead of t since the sample size is large enough.

For the 99% confidence interval the value of \alpha=1-0.99=0.01 and \alpha/2=0.005, with that value we can find the quantile required for the interval in the normal standard distribution.

z_{\alpha/2}=2.58

And replacing into the confidence interval formula we got:

12.8 - 2.58 \frac{16.6}{\sqrt{1003}}=11.448

12.8 + 2.58 \frac{16.6}{\sqrt{1003}} =14.152

And the 99% confidence interval would be given (11.448;14.152).

We are confident that about 11 to 14 are the number of books that the people had read on the last year on average, at 1% of significance.

3 0
3 years ago
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