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2 years ago

If 2a:3b=5:6 and 3b:2c=36:15,then find a:c

Mathematics

1 answer:

Salsk061 [2.6K]2 years ago

7 0

Answer:

a:c=2.5:1.25

Step-by-step explanation:

2a:3b=5:6

2(2.5):3(2)=5:6

3(2):2(1.25)=6:2.5=36:15

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Suppose the true proportion of students at a college who study abroad is 0.25. I select a random sample of 40 students from the

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Answer:

$\mu_{\hat p} = 0.25$

$\sigma_{\hat p}= \sqrt{\frac{0.25*(1-0.25)}{40}}= 0.0685$

$z = \frac{0.2-0.25}{0.0685}= -0.730$

And we can use the normal standard distribution or excel to find this probability and we got:

$P(\hat p$

Step-by-step explanation:

We define the parameter as the proportion of students at a college who study abroad and this value is known $p =0.25$ , we select a sample size of n =40 and we are interested in the probability associated to the sample proportion, but we know that the distirbution for the sample proportion is given by:

$\hat p \sim N(p , \sqrt{\frac{p(1-p)}{n}}$

And the paramters for this case are:

$\mu_{\hat p} = 0.25$

$\sigma_{\hat p}= \sqrt{\frac{0.25*(1-0.25)}{40}}= 0.0685$

We want to find the following probability:

$P(\hat p< 0.2)$

For this case since we know the distribution for the sample proportion we can use the z score formula given by:

$z = \frac{\hat p -\mu_{\hat p}}{\sigma_{\hat p}}$

Replacing the info given we got:

$z = \frac{0.2-0.25}{0.0685}= -0.730$

And we can use the normal standard distribution or excel to find this probability and we got:

$P(\hat p$

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The heights of a certain type of tree are approximately normally distributed with a mean height p = 5 ft and a standard

arsen [322]

Answer:

A tree with a height of 6.2 ft is 3 standard deviations above the mean

Step-by-step explanation:

⇒ $1^s^t$ statement: A tree with a height of 5.4 ft is 1 standard deviation below the mean(FALSE)

an X value is found Z standard deviations from the mean mu if:

$\frac{X-\mu}{\sigma} = Z$

In this case we have: $\mu=5\ ft$ $\sigma=0.4\ ft$

We have four different values of X and we must calculate the Z-score for each

For X =5.4\ ft

$Z=\frac{X-\mu}{\sigma}\\Z=\frac{5.4-5}{0.4}=1$

Therefore, A tree with a height of 5.4 ft is 1 standard deviation above the mean.

⇒ $2^n^d$ statement:A tree with a height of 4.6 ft is 1 standard deviation above the mean. (FALSE)

For X =4.6 ft

$Z=\frac{X-\mu}{\sigma}\\Z=\frac{4.6-5}{0.4}=-1$

Therefore, a tree with a height of 4.6 ft is 1 standard deviation below the mean .

⇒ $3^r^d$ statement:A tree with a height of 5.8 ft is 2.5 standard deviations above the mean (FALSE)

For X =5.8 ft

$Z=\frac{X-\mu}{\sigma}\\Z=\frac{5.8-5}{0.4}=2$

Therefore, a tree with a height of 5.8 ft is 2 standard deviation above the mean.

⇒ $4^t^h$ statement:A tree with a height of 6.2 ft is 3 standard deviations above the mean. (TRUE)

For X =6.2\ ft

$Z=\frac{X-\mu}{\sigma}\\Z=\frac{6.2-5}{0.4}=3$

Therefore, a tree with a height of 6.2 ft is 3 standard deviations above the mean.

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3 years ago

If 2a:3b=5:6 and 3b:2c=36:15,then find a:c​

If 2a:3b=5:6 and 3b:2c=36:15,then find a:c