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2 years ago

What is the ratio of 60:48

Mathematics

2 answers:

n200080 [17]2 years ago

7 0

Answer:

5:4

Step-by-step explanation:

60/48

=> Dividing numerator and denominator by 12,

=> 5/4 = 5:4

wlad13 [49]2 years ago

4 0

Answer:

$\pmb{\sf{5:4 }}$

Step-by-step explanation:

The ratio of 60:48

${\sf{\dfrac{60}{48} }}$

${\sf{\dfrac{10}{8} }}$

${\sf{\dfrac{5}{4}}}$

${\sf{5:4 }}$

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Blood pressure: High blood pressure has been identified as a risk factor for heart attacks and strokes. The proportion of U.S. a

Y_Kistochka [10]

Answer:

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

For this case we know this:

$n=37 , p=0.2$

We can find the standard error like this:

$SE = \sqrt{\frac{\hat p (1-\hat p)}{n}}= \sqrt{\frac{0.2*0.8}{37}}= 0.0658$

So then our random variable can be described as:

$p \sim N(0.2, 0.0658)$

Let's suppose that the question on this case is find the probability that the population proportion would be higher than 0.4:

$P(p>0.4)$

We can use the z score given by:

$z = \frac{p -\mu_p}{SE_p}$

And using this we got this:

$P(p>0.4) = 1-P(z< \frac{0.4-0.2}{0.0658}) = 1-P(z$

And we can find this probability using the Ti 84 on this way:

2nd> VARS> DISTR > normalcdf

And the code that we need to use for this case would be:

1-normalcdf(-1000, 3.04; 0;1)

Or equivalently we can use:

1-normalcdf(-1000, 0.4; 0.2;0.0658)

Step-by-step explanation:

We need to check if we can use the normal approximation:

$np = 37 *0.2 = 7.4 \geq 5$

$n(1-p) = 37*0.8 = 29.6\geq 5$

We assume independence on each event and a random sampling method so we can conclude that we can use the normal approximation and then ,the population proportion have the following distribution :

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

For this case we know this:

$n=37 , p=0.2$

We can find the standard error like this:

$SE = \sqrt{\frac{\hat p (1-\hat p)}{n}}= \sqrt{\frac{0.2*0.8}{37}}= 0.0658$

So then our random variable can be described as:

$p \sim N(0.2, 0.0658)$

Let's suppose that the question on this case is find the probability that the population proportion would be higher than 0.4:

$P(p>0.4)$

We can use the z score given by:

$z = \frac{p -\mu_p}{SE_p}$

And using this we got this:

$P(p>0.4) = 1-P(z< \frac{0.4-0.2}{0.0658}) = 1-P(z$

And we can find this probability using the Ti 84 on this way:

2nd> VARS> DISTR > normalcdf

And the code that we need to use for this case would be:

1-normalcdf(-1000, 3.04; 0;1)

Or equivalently we can use:

1-normalcdf(-1000, 0.4; 0.2;0.0658)

7 0

2 years ago

Fifty-eight percent of the fish in a large pond are minnows. Imagine scooping out a simple random sample of 20 fish from the pon

Makovka662 [10]

Answer:

The standard deviation is 0.1104. The 10% condition is met because it is very likely there are more than 200 minnows in the pond.

Step-by-step explanation:

The standard deviation for a proportion is:

σ = √(pq/n)

where p is the proportion, q is 1−p, and n is the sample size.

σ = √(0.58 × 0.42 / 20)

σ = 0.1104

Since the pond is large, there's probably more than 200 minnows, so the 10% condition is met.

7 0

3 years ago

New York City is one of the most expensive cities in the US for lodging. The mean hotel room rate is $244.00 per night; assume t

Illusion [34]

Answer:

22.96% probability that a hotel room costs between $250.00 and $285.00

Step-by-step explanation:

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

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

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

$\mu = 244, \sigma = 55$