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2 years ago

Ralph is 3 times as old as Sara. In 6 years , Ralph will be only twice as old as Sara will be then. Find Ralph’s age now .

Mathematics

1 answer:

Delvig [45]2 years ago

3 0

Answer:

Ralph's current age is 18.

Step-by-step explanation:

Let r and s represent the current ages of Ralph and Sara respectively. Our task here is to determine r, Ralph's age now.

If Ralph is 3 times as old as Sara now, then r = 3s.

Six years from now, Ralph's age will be r + 6 and Sara's will be s + 6. Ralph will be only twice as old as Sara will be then. This can be represented algebraically as

r + 6 = 2(s + 6).

We now have the following system of linear equations to solve:

r + 6 = 2s + 12, or r - 2s = 6, and r = 3s (found earlier, see above).

r - 2s = 6

r = 3s

Substituting 3s for r in r - 2s = 6, we get 3s = 2s + 6, or s = 6. Sara is 6 years old now, meaning that Ralph is 3(6 years), or 18 years old.

Ralph's current age is 18.

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The average yield of the standard variety off soybeans per acre on a farm in a region is 48.8 bushels per acre, with a standard

I am Lyosha [343]

Answer:

$z=\frac{53.4-48.8}{\frac{12}{\sqrt{36}}}=2.3$

$p_v =P(Z>2.3)=1-P(Z$

If we compare the p value and the significance level given $\alpha=0.01$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, so we can conclude that the true mean is not significantly higher than 48.8 at 1% of signficance.

Step-by-step explanation:

1) Data given and notation

$\bar X=53.4$ represent the sample mean

$\sigma=12$ represent the population standard deviation assumed

$n=36$ sample size

$\mu_o =48.8$ represent the value that we want to test

$\alpha=0.01$ represent the significance level for the hypothesis test.

z would represent the statistic (variable of interest)

$p_v$ represent the p value for the test (variable of interest)

2) State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the true mean is higher than 48.8, the system of hypothesis would be:

Null hypothesis: $\mu \leq 48$

Alternative hypothesis: $\mu > 48$

Since we assume that know the population deviation, is better apply a z test to compare the actual mean to the reference value, and the statistic is given by:

$z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}}$ (1)

z-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".

3) Calculate the statistic

We can replace in formula (1) the info given like this:

$z=\frac{53.4-48.8}{\frac{12}{\sqrt{36}}}=2.3$

4)P-value

Since is a right tailed test the p value would be:

$p_v =P(Z>2.3)=1-P(Z$

5) Conclusion

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3 years ago

10+3+(2×0.1)+(3×0.01)+(5×0.001) in word form?

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Simply add and multiply the numbers and you get 13.235
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3 years ago

A sample size 25 is picked up at random from a population which is normally

Margarita [4]

Answer:

a) P(X < 99) = 0.2033.

b) P(98 < X < 100) = 0.4525

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the z-score 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 p-value, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

Mean of 100 and variance of 36.

This means that $\mu = 100, \sigma = \sqrt{36} = 6$

Sample of 25:

This means that $n = 25, s = \frac{6}{\sqrt{25}} = 1.2$

(a) P(X<99)

This is the pvalue of Z when X = 99. So

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

By the Central Limit Theorem

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

$Z = \frac{99 - 100}{1.2}$

$Z = -0.83$

$Z = -0.83$ has a pvalue of 0.2033. So

P(X < 99) = 0.2033.

b) P(98 < X < 100)

This is the pvalue of Z when X = 100 subtracted by the pvalue of Z when X = 98. So

X = 100

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

$Z = \frac{100 - 100}{1.2}$

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X = 98

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

$Z = \frac{98 - 100}{1.2}$

$Z = -1.67$

$Z = -1.67$ has a pvalue of 0.0475

0.5 - 0.0475 = 0.4525

P(98 < X < 100) = 0.4525

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