What shape are possible cross section cylinder?

This is incorrect. Divide 378/60 and the quotient is the number of hours while the remainder is the minutes. In this case, it is 6 hours and 18 minutes. Please mark Brainliest!!!

5 0

3 years ago

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Adnan states if cos30°≈0.866, then sin30°≈0.866.

kiruha [24]

Answer:

Adnan is incorrect because cosx and sin(90 - x) are always equivalent in any right triangle ⇒ first answer

Step-by-step explanation:

In any right triangle there are two acute angles sum of them is 90°

∴ If one of them x°, then the other is (90 - x)°

∵ cos x = adjacent/hypotenuse

∵ sin(90 - x) = opposite/hypotenuse

∵ The opposite of ∠(90 - x) is the adjacent of ∠x

∴ cosx = sin(90 - x)

8 0

3 years ago

Bodytemperatures ofhealthy koalasarenormally distributed with a mean of 35.6°C and a standard deviation of 1.3°C. a.What is the

Licemer1 [7]

Answer:

a

$P(X < 35) = 32.3 \%$

b

$P(\= X < 35) = 0.6 \%$

Here the probability of koalas mean temperature being less than 35 °C is very small hence the koalas are not healthy

c

A potential confounding variable for this study is the population of the koalas because in the first question the population was not taken into account and the probability was $P(X < 35) = 32.3 \%$ but when the population was taken into account (i.e n = 30) the probability became

$P(\= X < 35) = 0.6 \%$

Step-by-step explanation:

From the question we are told that

The mean is $\mu = 35.6^oC$

The standard deviation is $s = 1.3^oC$

The sample size is n = 30

Generally the probability that a health koala has a body temperature less than 35.0°C is mathematically represented as

$P(X < 35) = P(\frac{X - \mu }{s} < \frac{35 - 35.6}{1.3} )$

Here $(\frac{X - \mu }{s} = Z (The \ standardized \ value \ of \ X )$

So

$P(X < 35) = P(Z < -0.46)$

From the z-table P(Z < -0.46) = 0.323

So

$P(X < 35) = 0.323$

Converting to percentage

$P(X < 35) = 0.323 * 100$

$P(X < 35) = 32.3 \%$

considering question b

The sample mean is $\= x = 35$

Generally the standard error of the mean is mathematically represented as

$\sigma_{\= x} = \frac{s}{\sqrt{n} }$

=> $\sigma_{\= x} = \frac{1.3}{\sqrt{30} }$

=> $\sigma_{\= x} = 0.2373$

Generally the probability of the mean body temperature of koalas being less than 35.0°C is mathematically represented as

$P(\= X < 35) = P(\frac{\= X - \mu }{\sigma_{\= x }} < \frac{35 -35.6}{0.2373 } )$

$P(\= X < 35) = P(Z< -2.53 )$

From the z-table we have that

$P(Z< -2.53 ) = 0.006$

So

$P(\= X < 35) = 0.006 /tex] Converting to percentage [tex]P(\= X < 35) = 0.006 * 100$

$P(\= X < 35) = 0.6 \%$

Here the probability of koalas mean temperature being less than 35 °C is very small hence the koalas are not healthy

A potential confounding variable for this study is the population of the koalas because in the first question the population was not taken into account and the probability was $P(X < 35) = 32.3 \%$ but when the population was taken into account (i.e n = 30) the probability became

$P(\= X < 35) = 0.6 \%$

4 0

4 years ago