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3 years ago

If you solve using the equation for the constant of proportionality, y/x = k with k = 4/5 and y = 2/3, what is the value of x?

Mathematics

2 answers:

ruslelena [56]3 years ago

6 0

Answer:

x=5/6

Step-by-step explanation:

k=4/5

3 and 5 can go into 15

10/15 and 12/15

10/15 divided by x =12/15

x=5/6

GenaCL600 [577]3 years ago

3 0

Answer:

5/6

Step-by-step explanation:

The given equation is $\frac{y}{x}=k$

Solve the equation for x

Cross multiplying, we get

$y=xk$

Divide both sides by k

$x=\frac{y}{k}$

Substituting the given values of y and k

$x=\frac{2/3}{4/5}$

Flip the denominator

$x=\frac{2}{3}\cdot\frac{5}{4}\\\\x=\frac{5}{6}$

Thus, the value of x is 5/6

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A simple random sample of 20 days in which Parsnip ate seeds was selected, and the mean amount of time it took him to eat the se

bearhunter [10]

Answer:

$(14.5-13.9) -1.69 \sqrt{\frac{3.98^2}{20}+ \frac{4.03^2}{17}} = -1.63$

$(14.5-13.9) +1.69 \sqrt{\frac{3.98^2}{20}+ \frac{4.03^2}{17}} = 2.83$

And the confidence interval for this case $-1.63 \leq \mu_1 -\mu_2 \leq 2.83$

Step-by-step explanation:

We know the following info from the problem

$\bar X_1 = 14.5$ sample mean for the group 1

$s_1 = 3.98$ the standard deviation for the group 1

$n_1= 20$ the sample size for group 1

$\bar X_2 = 13.9$ sample mean for the group 2

$s_1 = 4.03$ the standard deviation for the group 2

$n_2= 17$ the sample size for group 2

We have all the conditions satisifed since we have random samples.

We want to construct a confidence interval for the true difference of means and the correct formula for this case is:

$(\bar X_1 -\bar X_2) \pm t_{\alpha/2}\sqrt{\frac{s^2_1}{n_1} +\frac{s^2_2}{n_2}}$

The degrees of freedom are given :

$df = n_1 +n_2- 2 = 20+17-2=35$

The confidence level is 0.9 or 90% and the significance level is $\alpha=1-0.9=0.1$ and $\alpha/2 = 0.05$ and the critical value for this case is:

$t_{\alpha/2} = 1.69$

And replacing the info given we got:

$(14.5-13.9) -1.69 \sqrt{\frac{3.98^2}{20}+ \frac{4.03^2}{17}} = -1.63$

$(14.5-13.9) +1.69 \sqrt{\frac{3.98^2}{20}+ \frac{4.03^2}{17}} = 2.83$

And the confidence interval for this case $-1.63 \leq \mu_1 -\mu_2 \leq 2.83$

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Hope this helped :]

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