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2 years ago

8

What is the probability that a randomly chosen number between 1 and 100 is divisible by 3, given that the number has at least on

e digit equal to 5

1 answer:

ioda2 years ago

6 0

Answer:

Step-by-step explanation:

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Divide R300 000 between Leona and Janet in the ratio 3:2.

ella [17]

Answer:

<em><u>let </u></em><em><u>the </u></em><em><u>ratio</u></em><em><u> be</u></em><em><u> in</u></em><em><u> </u></em><em><u>x </u></em>

<em><u>a/</u></em><em><u>q</u></em><em><u>. </u></em><em><u>3</u></em><em><u>x</u></em><em><u> </u></em><em><u>+</u></em><em><u> </u></em><em><u>2</u></em><em><u>x</u></em><em><u> </u></em><em><u>=</u></em><em><u> </u></em><em><u>3</u></em><em><u>0</u></em><em><u>0</u></em><em><u>0</u></em><em><u>0</u></em><em><u>0</u></em>

<em><u> </u></em><em><u> </u></em><em><u>5</u></em><em><u>x</u></em><em><u> </u></em><em><u>=</u></em><em><u> </u></em><em><u>3</u></em><em><u>0</u></em><em><u>0</u></em><em><u>0</u></em><em><u>0</u></em><em><u>0</u></em>

<em><u>x </u></em><em><u>=</u></em><em><u> </u></em><em><u>3</u></em><em><u>0</u></em><em><u>0</u></em><em><u>0</u></em><em><u>0</u></em><em><u>0</u></em><em><u>/</u></em><em><u>5</u></em>

<em><u>x=</u></em><em><u> </u></em><em><u>6</u></em><em><u>0</u></em><em><u>0</u></em><em><u>0</u></em><em><u>0</u></em>

<em><u>Leona </u></em><em><u>=</u></em><em><u> </u></em><em><u>6</u></em><em><u>0</u></em><em><u>0</u></em><em><u>0</u></em><em><u>0</u></em><em><u>*</u></em><em><u>3</u></em><em><u> </u></em><em><u>=</u></em><em><u> </u></em><em><u>1</u></em><em><u>8</u></em><em><u>0</u></em><em><u>0</u></em><em><u>0</u></em><em><u>0</u></em>

<em><u>Janet </u></em><em><u>=</u></em><em><u> </u></em><em><u>6</u></em><em><u>0</u></em><em><u>0</u></em><em><u>0</u></em><em><u>0</u></em><em><u>*</u></em><em><u>2</u></em><em><u> </u></em><em><u>=</u></em><em><u> </u></em><em><u>1</u></em><em><u>2</u></em><em><u>0</u></em><em><u>0</u></em><em><u>0</u></em><em><u>0</u></em>

<em><u>hope</u></em><em><u> it</u></em><em><u> helps</u></em>

5 0

1 year ago

Find the sum of this infinite geometric series where a1=0.3 and r=0.1

ratelena [41]

Answer:

1/3

Step-by-step explanation:

The formula for computing the sum of an infinite geometric series is

$S=\frac{a_1}{1-r}$ where r is between -1 and 1 and $r$ is the common ratio, and $a_1$ is the first term of the series.

So let's plug in:

$S=\frac{0.3}{1-0.1}$

$S=\frac{0.3}{0.9}$

$S=\frac{3}{9}$ I multiplied bottom and top by 10.

$S=\frac{1}{3}$ I divided top and bottom by 3.

The sum is 1/3.

8 0

2 years ago

HELP ALMOST CLASS TIME

Lesechka [4]

Answer:

B

Step-by-step explanation:

4 0

2 years ago

Read 2 more answers

100 points! Will mark brainliest! Only answer if u know it. Random answers will be reported

Setler79 [48]

Step-by-step explanation:

Hey there!

According to the question, you need to find out the distance from your friend's house to school.

<em>Firstly figure out the coordinate of your school and school. Then use formula of distance.</em>

So, we find the coordinates as (7,7) and (5,0).

So, x 1 = 7 y1= 7

x 2 = 5 y2 = 0

~ Use distance formula.

$d = \sqrt{( {x2 - x1)}^{2} + ( {y2 - y1)}^{2} }$

~ Put all values.

$d = \sqrt{ {(5 - 7)}^{2} + ( {0 - 7)}^{2} }$

~ Simplify it.

$d = \sqrt{ {( - 2)}^{2} + {( - 7)}^{2} }$

$d = \sqrt{4 + 49}$

$d = \sqrt{53}$

D= 7.280

Therefore, the distance between your friend's house and your school is 7.3 units.

<em><u>Hope</u></em><em><u> </u></em><em><u>it</u></em><em><u> helps</u></em><em><u>.</u></em><em><u>.</u></em><em><u>.</u></em><em><u>.</u></em>

8 0

2 years ago

Read 2 more answers

Could someone help?<br> Thank you

frozen [14]

Answer:

See Below

Step-by-step explanation:

Thankfully, all the parts you need are given.

First, find the surface of one triangle

A = 1/2bh

A = 1/2(15)(9)

A = 1/2(135)

A = 67.5

Multiply by 4 to get the sum of all the sides.

67.5(4) = 270

Now find the surface area of one square

A = bh

A = (15)(15)

A = 225

multiply by 5 to get all 5 sides

225(5) = 1125

Add sums together

1125 + 270 = 1395

3 0

2 years ago