If y varies inversely with x and y= 17 when z = 11 when y = 11 find y when x=19 .

Mathematics

1 answer:

BaLLatris [955]3 years ago

8 0

Where’s your z? incomplete question

You might be interested in

What’s 1+1? Is it 929? Uwu

Allisa [31]

Answer:

No it's 2 Lol not 929 UwU

3 0

3 years ago

Consider the following set of data:

Marrrta [24]

Mean is all the data added up and divided by how many numbers there are
Median is when you sort the numbers in order from smallest to largest and the one in the middle, if there are two in the middle find the average
Mode is number ir numbers that appears the most
Tell me if you want the answer

4 0

3 years ago

What is the median of the data set?

Margaret [11]

22 - Median is in the middle

8 0

3 years ago

Read 2 more answers

Assume ∆ EFG ≅ ∆ JKL. If m∠J =90, m∠K =26, what is the measure of ∠E ?

Anna11 [10]

Answer:

90 A pex

Step-by-step explanation:

8 0

3 years ago

A random sample of 31 charge sales showed a sample standard deviation of $50. a 90% confidence interval estimate of the populati

Marysya12 [62]

The $100(1-\alpha)\%$ confidence interval of a standard deviation is given by:

$\sqrt{ \frac{(n-1)s^2}{\chi^2_{1- \frac{\alpha}{2} } }} \leq\sigma\leq\sqrt{ \frac{(n-1)s^2}{\chi^2_{\frac{\alpha}{2} } }}$

Given a sample size of 31 charge sales, the degree of freedom is 30 and for 90% confidence interval,

$\chi^2_{1- \frac{\alpha}{2} }=43.773 \\ \\ \chi^2_{\frac{\alpha}{2} }=18.493$

Therefore, the 90% confidence interval for the standard deviation is given by

$\sqrt{ \frac{(31-1)50^2}{43.773 }} \leq\sigma\leq\sqrt{ \frac{(31-1)50^2}{18.493 }} \\ \\ \Rightarrow\sqrt{ \frac{(30)2500}{43.773 }} \leq\sigma\leq\sqrt{ \frac{(30)2500}{18.493 }} \\ \\ \Rightarrow\sqrt{ \frac{75000}{43.773 }} \leq\sigma\leq\sqrt{ \frac{75000}{18.493 }} \\ \\ \Rightarrow \sqrt{1713.38} \leq\sigma\leq \sqrt{4055.59} \\ \\ \Rightarrow41.4\leq\sigma\leq63.7$

3 0

3 years ago