Can someone help with this please

The ratio of the current ages of two RSM summer camp students Dylan and Mike is 3:4. If Dylan had been 6 years younger and Mike

My name is Ann [436]

Answer:

13:16, 3:4, 5:8.

Step-by-step explanation:

The ratio of the current ages of two students Dylan and Mike is 3:4.

Let their ages are 3x and 4x respectively.

If Dylan had been 6 years younger and Mike 6 years older, the age of Mike would have been 6 times the age of Dylan.

Hence, we can write, 6(3x - 6) = (4x + 6)

⇒ 18x - 4x = 36 + 6

⇒14x = 42

⇒ x = 3

Therefore, the current age of Dylan is 3×3 =9 years and that of Mike is 3×4 =12 years.

Therefore, the ratio of Dylan's current age to Mike's current age is 3:4. The ratio of their ages in four years ago will be (9-4):(12-4) =5:8

The ratio of their ages in four years from now will be (9+4):(12+4) =13:16

Hence, the ratios in descending order is 13:16, 3:4, 5:8.

(Answer)

6 0

3 years ago

A recent survey of 50 executives who were laid off during a recent recession revealed it took a mean of 26 weeks for them to fin

drek231 [11]

Answer:

The pvalue of 0.0113 < 0.05 means that there is sufficient evidence to conclude that the mean time to find another position is less than 28 weeks at the 5% level of significance

Step-by-step explanation:

The null hypothesis is:

$H_{0} = 28$

The alternate hypotesis is:

$H_{1} < 28$

The test statistic is:

$z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}$

In which X is the sample mean, $\mu$ is the value tested at the null hypothesis, $\sigma$ is the standard deviation and n is the size of the sample.

A recent survey of 50 executives who were laid off during a recent recession revealed it took a mean of 26 weeks for them to find another position.

This means that $n = 50, X = 26$

Assume the population standard deviation is 6.2 weeks.

This means that $\sigma = 6.2$

Does the data provide sufficient evidence to conclude that the mean time to find another position is less than 28 weeks at the 5% level of significance

We have to find the pvalue of Z, looking at the z-table, when $\mu = 28$ . It if is lower than 0.05, it provides evidence.

$z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}$

$z = \frac{26 - 28}{\frac{6.2}{\sqrt{50}}}$

$z = -2.28$

$z = -2.28$ has a pvalue of 0.0113 < 0.05.

The pvalue of 0.0113 < 0.05 means that there is sufficient evidence to conclude that the mean time to find another position is less than 28 weeks at the 5% level of significance

5 0

3 years ago

Define Interquartile Range.<br><br> I will check for plagiarism.

nlexa [21]

Interquartile range is the range is the number in the dead center, you have to divide the number line into 2 sections. The middle of everything and the middle of both section is the interquartile range

Hope this helps!

5 0

3 years ago

Read 2 more answers

Write 1/3 as a power with an integer base

Viefleur [7K]

Answer:

$3^{-1}$