In triangle PQR, line PO bisects

Mathematics

1 answer:

Valentin [98]3 years ago

5 0

15.

To find this, here is the work in the photo attached:

You might be interested in

How does the sample size affect the validity of an empirical argument? A. The larger the sample size the better. B. The smaller

astra-53 [7]

Answer:

A. The larger the sample size the better.

Step-by-step explanation:

Central Limit Theorem

The Central Limit Theorem establishes 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.

For a proportion p in a sample of size n, the sampling distribution of the sample proportion will be approximately normal with mean $\mu = p$ and standard deviation $s = \sqrt{\frac{p(1-p)}{n}}$

In this question:

We have to look at the standard error, which is:

$s = \frac{\sigma}{\sqrt{n}}$

This means that an increase in the sample size reduces the standard error, and thus, the larger the sample size the better, and the correct answer is given by option a.

7 0

3 years ago

If the probability is 0.54 that Stock A will increase in value during the next month and the probability is 0.68 that Stock B wi

Aloiza [94]

P(A) =0.54

P(B)= 0.68

P'(A)= 1-0.54 = 0.46

P'(B)= 1- 0.68 = 0.32

The probability of neither of both event will occur:

= P'(A)×P'(B)

=0.46 × 0.32

=0.1472

3 0

3 years ago

Will someone help with this ! I was allowed a retry for it and I don’t know how to do it

pashok25 [27]

Answer:

$\frac{dN}{dt}=k(725-N)\\separating the variables and integrating\\\int \frac{dN}{725-N}=\int kdt+c\\-log (725-N)=kt+c\\log (725-N)=-kt-c\\(725-N)=e^{-kt-c} =e^{-kt} *e^{-c} =Ce^{-kt} \\when t=0,N=400\\725-400=Ce^{0} =C\\C=325\\when t=3,N=650\\725-650=325e^{-3k} \\\frac{75}{325}=(e^{-k})^3\\\\e^{-k} =(\frac{3}{13}) ^{\frac{1}{3} } \\when t=5\\725-N=325(\frac{3}{13}) ^{\frac{5}{3} } =325*\frac{3}{13}*(\frac{3}{13} )^{\frac{2}{3}}\\N=725-75*(\frac{3}{13} )^{\frac{2}{3} }=725-28=697$