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3 years ago

The diameter of a circle measures 10 cm. What is the circumference of the circle?

Mathematics

1 answer:

Klio2033 [76]3 years ago

8 0

Answer:

circumference of a circle=πd where d is the diameter of the circle.

this implies, the circumference=3.14*10cm

=31.4cm

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Deposits $10 in an account that earns an interest rate of 5% compounded annually. What is the balance in the account at the end

d1i1m1o1n [39]

<span>Compound Interest is calculated on the initial payment and also on the interest of previous periods.

So after 1 year w/ 5% interest, $10 -> $10*1.05 = $10.5
after 2 year, the total becomes $10.5*1.05 = $11.025

formula to calculate is 10*(1.05)^50 = $114.67

ans is A</span>

8 0

3 years ago

Sean bought a laptop and a printer. He sold both items for £1,500 getting £1,350 just for the laptop. Sean made a 24% profit on

vitfil [10]

Answer:

Step-by-step explanation:

Sean purchased a printer and laptop

He sold both for 1,500

He sold the laptop at 1,350

Therefore the cost of the printer can be calculated as follows

1500-1350

= 150

24/100×150

= 0.24×150

= 36

Henve the cost of the printer is £36

4 0

3 years ago

Calculate the sample mean and sample variance for the following frequency distribution of heart rates for a sample of American a

spin [16.1K]

Answer:

$Mean = 68.9$

$s^2 =18.1$ --- Variance

Step-by-step explanation:

Given

$\begin{array}{cccccc}{Class} & {51-58} & {59-66} & {67-74} & {75-82} & {83-90} \ \\ {Frequency} & {6} & {3} & {11} & {13} & {4} \ \end{array}$

Solving (a): Calculate the mean.

The given data is a grouped data. So, first we calculate the class midpoint (x)

For 51 - 58.

$x = \frac{1}{2}(51+58) = \frac{1}{2}(109) = 54.5$

For 59 - 66

$x = \frac{1}{2}(59+66) = \frac{1}{2}(125) = 62.5$

For 67 - 74

$x = \frac{1}{2}(67+74) = \frac{1}{2}(141) = 70.5$

For 75 - 82

$x = \frac{1}{2}(75+82) = \frac{1}{2}(157) = 78.5$

For 83 - 90

$x = \frac{1}{2}(83+90) = \frac{1}{2}(173) = 86.5$

So, the table becomes:

$\begin{array}{cccccc}{x} & {54.5} & {62.5} & {70.5} & {78.5} & {86.5} \ \\ {Frequency} & {6} & {3} & {11} & {13} & {4} \ \end{array}$

The mean is then calculated as:

$Mean = \frac{\sum fx}{\sum f}$

$Mean = \frac{54.5*4+62.5*3+70.5*11+78.5*13+86.5*4}{6+3+11+13+4}$

$Mean = \frac{2547.5}{37}$

$Mean = 68.9$ -- approximated

Solving (b): The sample variance:

This is calculated as:

$s^2 =\frac{\sum (x - \overline x)^2}{\sum f - 1}$

So, we have:

$s^2 =\frac{(54.5-68.9)^2+(62.5-68.9)^2+(70.5-68.9)^2+(78.5-68.9)^2+(86.5-68.9)^2}{37 - 1}$

$s^2 =\frac{652.8}{36}$

$s^2 =18.1$ -- approximated

5 0

3 years ago

Select the correct answer. Simplify the expression.<br> (see the screenshot)

Sliva [168]

Answer:

Option D

Step-by-step explanation:

Given expression has been given as,

$\sqrt[5]{224x^{11}y^8}$

$\sqrt[5]{224x^{11}y^8}=\sqrt[5]{2\times 2\times 2\times 2\times 2\times 7(x^{11})(y^8)}$

$=\sqrt[5]{(2^5)\times (7)(x^{10}\times x)(y^5\times y^3)}$

$=2^{\frac{5}{5}}\times 7^{\frac{1}{5}}\times x^{\frac{10}{5}}\times x^{\frac{1}{5}}\times y^{\frac{5}{5} }\times y^{\frac{3}{5} }$

$=2\times 7^{\frac{1}{5}}\times x^2\times y\times x^{\frac{1}{5} }\times y^{\frac{3}{5} }$

$=2x^2y\sqrt[5]{7xy^3}$

Option D will be the answer.

7 0

3 years ago

I don’t understand this

Alexandra [31]

Answer:

-15.41

Step-by-step explanation:

big brain?

5 0

3 years ago

Read 2 more answers