All categories

3 years ago

Determine the radius of a circle given by the equation x^2+y^2=50.

Mathematics

1 answer:

bogdanovich [222]3 years ago

5 0

Answer:

r = 5√2

Step-by-step explanation:

Circle Formula: (x - h)² + (y - k)² = r²

(h, k) is center, <em>r </em>is radius.

Since we are given 50 as r², we set equal and solve:

r² = 50

r = √50

r = 5√2

You might be interested in

Audrey deposited $40,000 in a savings account with simple interest. Five years later, she had earned $30,000 in interest. What w

Fittoniya [83]

$~~~~~~ \textit{Simple Interest Earned} \\\\ I = Prt\qquad \begin{cases} I=\textit{interest earned}\dotfill&\$30000\\ P=\textit{original amount deposited}\dotfill & \$40000\\ r=rate\to r\%\to \frac{r}{100}\\ t=years\dotfill &5 \end{cases} \\\\\\ 30000=(40000)(\stackrel{\%}{\frac{r}{100}})(5)\implies \cfrac{30000}{(40000)(5)}=\cfrac{r}{100}\implies \cfrac{3}{20}=\cfrac{r}{100} \\\\\\ \cfrac{300}{20}=r\implies \stackrel{\%}{15}=r$

4 0

2 years ago

Help:(, i’m not sure how to do this

Schach [20]

Answer:

i think you have to multiply the sides..

Step-by-step explanation:

5 0

2 years ago

Read 2 more answers

Mr. McLaughlin was setting up for parent teacher meetings in the gym. The door was 42 inches wide and 84 inches tall. He needed

saul85 [17]

The table would fit because if he rolled the table in which sideways it would fit since its smaller than the door

5 0

3 years ago

For parts a and bâ, use technology to estimate the following. âa) The critical value of t for a 90â% confidence interval with df

rjkz [21]

Answer:

a) For the 90% confidence interval the value of $\alpha=1-0.9=0.1$ and $\alpha/2=0.05$ , with that value we can find the quantile required for the interval in the t distribution with df =3. And we can use the folloiwng excel code: "=T.INV(0.05,3)" and we got:

$t_{\alpha/2} =\pm 2.35$

b) For the 99% confidence interval the value of $\alpha=1-0.99=0.01$ and $\alpha/2=0.005$ , with that value we can find the quantile required for the interval in the t distribution with df =106. And we can use the folloiwng excel code: "=T.INV(0.005,106)" and we got:

$t_{\alpha/2} =\pm 2.62$

Step-by-step explanation:

Previous concepts

The t distribution (Student’s t-distribution) is a "probability distribution that is used to estimate population parameters when the sample size is small (n<30) or when the population variance is unknown".

The shape of the t distribution is determined by its degrees of freedom and when the degrees of freedom increase the t distirbution becomes a normal distribution approximately.

The degrees of freedom represent "the number of independent observations in a set of data. For example if we estimate a mean score from a single sample, the number of independent observations would be equal to the sample size minus one."

Solution to the problem

Part a

For the 90% confidence interval the value of $\alpha=1-0.9=0.1$ and $\alpha/2=0.05$ , with that value we can find the quantile required for the interval in the t distribution with df =3. And we can use the folloiwng excel code: "=T.INV(0.05,3)" and we got:

$t_{\alpha/2} =\pm 2.35$

Part b

For the 99% confidence interval the value of $\alpha=1-0.99=0.01$ and $\alpha/2=0.005$ , with that value we can find the quantile required for the interval in the t distribution with df =106. And we can use the folloiwng excel code: "=T.INV(0.005,106)" and we got:

$t_{\alpha/2} =\pm 2.62$

3 0

3 years ago

Six donuts cost a total of $3.54 what is the cost for one donut? Six donuts cost a total of $3.54 what is the cost of one donut?

UkoKoshka [18]

Answer:

one doughnut costs 59 cents

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers